ML Aggarwal Class 12 Maths Solutions Section A Chapter 10 Probability MCQs

Utilizing ML Aggarwal Maths for Class 12 Solutions Chapter 10 Probability MCQs as a study aid can enhance exam preparation.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 10 Probability MCQs

Choose the correct answer from the given four options in questions (1 to 82) :

Question 1.
If A and B are two events and P (A’) = 0.3, P(B) = 0.4, P (A ∩ B’) = 0.5, then P (A ∪ B’) =
(a) 0.5
(b) 0.8
(c) 1
(d) 0.1
Answer:
(b) 0.8

Given P (A’) = 0.3
⇒ P(A) = 1 – P(A’)
= 1 – 0.3 = 0.7
∴ P (A ∪ B’) = (A) + P (B’) – P (A ∩ B’)
= 0.7 + (0.4) – 0.5
= 0.7 + 0.6 – 0.5
= 0.8

Question 2.
A die is thrown. If A is the event that the number obtained is greater than 3 and B is event that the number obtained is less than 5, then P (A ∪ B) is
(a) 1
(b) \(\frac{2}{5}\)
(c) \(\frac{3}{5}\)
(d) 0
Answer:
(c) \(\frac{3}{5}\)

A : event that the no. of obtained is > 3 = {4, 5, 6}
∴ n (A) = 3
B : event that the no. obtained is < 5 = {1, 2, 3, 4}

Question 3.
If P(A ∩ B) = \(\frac{1}{2}\) and P(A’ ∩ B’) = \(\frac{1}{3}\), P(A) = p and P(B) = 2p, then the value of p is
(a) \(\frac{1}{9}\)
(b) \(\frac{1}{3}\)
(c) \(\frac{4}{9}\)
(d) \(\frac{7}{18}\)
Answer:
(d) \(\frac{7}{18}\)

Given \(\frac{1}{3}\) = P(A’ ∩ B’)
= P(A ∪ B)’ = 1 – P(A ∪ B)
⇒ \(\frac{1}{3}\) = 1 – [P(A) + P(B) – P(A ∩ B)]
⇒ \(\frac{1}{3}\) = 1 – [p + 2p – \(\frac{1}{2}\)]
⇒ \(\frac{1}{3}\) = \(\frac{3}{2}\) – 3p
⇒ 3p = \(\frac{3}{2}-\frac{1}{3}=\frac{7}{6}\)
⇒ p = \(\frac{7}{18}\)

Question 4.
If P(A) = \(\frac{4}{5}\) and P(A ∩ B) = \(\frac{7}{10}\), then P(B/A) is equal to
(a) \(\frac{1}{10}\)
(b) \(\frac{1}{8}\)
(c) \(\frac{7}{8}\)
(d) \(\frac{17}{20}\)
Answer:
(c) \(\frac{7}{8}\)

P(B/A) = \(\frac{P(A \cap B)}{P(A)}\)
= \(\frac{\frac{7}{10}}{\frac{4}{5}}=\frac{7}{10} \times \frac{5}{4}=\frac{7}{8}\)

Question 5.
If A and B are two events such that P (A) = 0.2, P (B) = 0.4 and P (A ∪ B)=0.6, then P (A/B) is equal to
(a) 0
(b) 0.5
(c) 0.3
(0) 0.8
Answer:
(a) 0

∴ P(A ∩ B) = P(A) + P(B) – P(A ∪ B)
= 0.2 + 0.4 – 0.6
= 0
Then P (A/B) = \(\frac{P(A \cap B)}{P(B)}\) = 0

Question 6.
If P(A) = \(\frac{1}{2}\), P(B) = \(\frac{2}{5}\) and P (A ∪ B) = \(\frac{3}{5}\) then P (B/A) + P (A/B) =
(a) \(\frac{1}{4}\)
(b) \(\frac{1}{3}\)
(c) \(\frac{5}{12}\)
(d) \(\frac{7}{12}\)
Answer:
(d) \(\frac{7}{12}\)

We know that
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
⇒ \(\frac{3}{5}=\frac{3}{10}+\frac{2}{5}\) – P(A ∩ B)
⇒ P(A ∩ B) = \(\frac{3}{10}+\frac{2}{5}-\frac{3}{5}=\frac{3+4-6}{10}=\frac{1}{10}\)

∴ P(B/A) + P(A/B)
= \(\frac{P(A \cap B)}{P(A)}+\frac{P(A \cap B)}{P(B)}\)
= \(\frac{\frac{1}{10}}{\frac{3}{10}}+\frac{\frac{1}{10}}{\frac{2}{5}}=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)

Question 7.
If P(A) = 0.4, P(B) = 0.8 and P(B/A) = 0.6, then P(A ∪ B) is equal to
(a) 0.24
(b) 0.96
(c) 0.3
(d) 0.48
Answer:
(b) 0.96

Given P (A) = 0.4 ; P (B) = 0.8 and P (B/A) = 0.6
⇒ \(\frac{P(B \cap A)}{P(A)}\) = 0.6
⇒ P (A ∩ B) = 0.6 P (A) = 0.6 × 0.4 = 0.24
Thus
P (A ∪ B) = P (A) + P (B) – P (A ∩ B)
= 0.4 + 0.8 – 0.24
= 0.96

Question 8.
If P(A) = \(\frac{7}{13}\), P(B) = \(\frac{9}{13}\) and P(A ∩ B) = \(\frac{4}{13}\), then P(A’/B) is equal to
(a) \(\frac{6}{13}\)
(b) \(\frac{5}{13}\)
(c) \(\frac{4}{9}\)
(d) \(\frac{5}{9}\)
Answer:
(d) \(\frac{5}{9}\)

Given P(A) = \(\frac{7}{13}\), P(B) = \(\frac{9}{13}\)
and P(A ∩ B) = \(\frac{4}{13}\)
∴ P(A’/B) = \(\frac{\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}\right)}{\mathrm{P}(\mathrm{B})}\)
= \(\frac{P(B)-P(A \cap B)}{P(B)}\)
= \(\frac{\frac{9}{13}-\frac{4}{13}}{\frac{9}{13}}=\frac{\frac{5}{13}}{\frac{9}{13}}=\frac{5}{9}\)

Question 9.
If A and B are two events such that P(A) = \(\frac{1}{2}\) P(B) = \(\frac{1}{3}\) and P(A/B) = \(\frac{1}{4}\) then P(A’ ∩ B’) is equal to
(a) \(\frac{1}{12}\)
(b) \(\frac{3}{4}\)
(c) \(\frac{1}{4}\)
(d) \(\frac{3}{16}\)
Answer:
(c) \(\frac{1}{4}\)

Given P(A) = \(\frac{1}{2}\) P(B) = \(\frac{1}{3}\)
and P(A/B) = \(\frac{1}{4}\)
⇒ \(\frac{P(B \cap A)}{P(B)}=\frac{1}{4}\)
⇒ P(A ∩ B) = \(\frac{1}{4} \times \frac{1}{3}=\frac{1}{12}\)
∴ P(A’ ∩ B’) = P{(A ∪ B)’}
= 1 – P(A ∪ B)
= 1 – {P(A) + P(B) – P(A ∩ B)
= 1 – \(\left\{\frac{1}{2}+\frac{1}{3}-\frac{1}{12}\right\}\)
= 1 – \(\frac{6+4-1}{12}\)
= 1 – \(\frac{9}{12}=\frac{1}{4}\)

Question 10.
If A and B are events such that P (A) = 0.4, P (B) = 0.3 and P (A ∪ B) = 0.5, then P (B’ ∩ A) is equal to
(a) \(\frac{1}{2}\)
(b) \(\frac{1}{5}\)
(c) \(\frac{3}{10}\)
(d) \(\frac{2}{3}\)
Answer:
(b) \(\frac{1}{5}\)

We know that,
P (A ∪ B) = P (A) + P (B) – P (A ∩ B)
⇒ 0.5 = 0.4 + 0.3 – P (A n B)
P(A ∩ B) = 0.7 – 0.5 = 0.2
P (A ∩ B’) = P (A) – P (A ∩ B)
= 0.4 – 0.2
= 0.2
= \(\frac{1}{5}\)

Question 11.
If A and B are two events such that P(B) = \(\frac{3}{5}\), P(A/B) = \(\frac{1}{2}\) and P(A ∪ B) = \(\frac{4}{5}\), then P(A) equals
(a) \(\frac{3}{10}\)
(b) \(\frac{1}{5}\)
(c) \(\frac{1}{2}\)
(d) \(\frac{3}{5}\)
Answer:
(c) \(\frac{1}{2}\)

Given P(A/B) = \(\frac{1}{2}\)
⇒ \(\frac{P(A \cap B)}{P(B)}=\frac{1}{2}\)
⇒ P(A ∩ B) = \(\frac{1}{2} \times \frac{3}{5}=\frac{3}{10}\)

We know that,
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
⇒ \(\frac{4}{5}\) = P(A) + \(\frac{3}{5}-\frac{3}{10}\)
P(A) = \(\frac{4}{5}-\frac{3}{5}+\frac{3}{10}\)
= \(\frac{8-6+3}{10}=\frac{5}{10}\)
= \(\frac{1}{2}\)

Question 12.
If A and B are two events such that P (A) = 0.6, P (B) = 0.2 and P (A/B) = 0.5, then P (A’/B’) is equal to
(a) \(\frac{1}{10}\)
(b) \(\frac{3}{10}\)
(c) \(\frac{6}{7}\)
(d) \(\frac{3}{8}\)
Answer:
(d) \(\frac{3}{8}\)

Given P (A) = 0.6; P (B) = 0.2;
P(A/B) = 0.5

Question 13.
If A and B are two events such that P (A) > 0 and P (B) ≠ 1, then P (A’/B’) is equal to
(a) 1 – P (A/B)
(b) 1 – P (A’/B)
(c) \(\frac{1-\mathbf{P}(\mathbf{A} \cup \mathbf{B})}{\mathbf{P}\left(\mathbf{B}^{\prime}\right)}\)
(d) \(\frac{\mathbf{P}\left(\mathbf{A}^{\prime}\right)}{\mathbf{P}\left(\mathbf{B}^{\prime}\right)}\)
Answer:
(c) \(\frac{1-\mathbf{P}(\mathbf{A} \cup \mathbf{B})}{\mathbf{P}\left(\mathbf{B}^{\prime}\right)}\)

Since P (A) > 0 and P (B) ≠ 1
∴ P(A’/B’) = \(\frac{\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)}{\mathrm{P}\left(\mathrm{B}^{\prime}\right)}\)
= \(\frac{\mathrm{P}\left\{(\mathrm{A} \cup \mathrm{B})^{\prime}\right\}}{\mathrm{P}\left(\mathrm{B}^{\prime}\right)}\)
= \(\frac{1-P(A \cup B)}{P\left(B^{\prime}\right)}\)

Question 14.
If A and B are two events such that P(A) = \(\frac{3}{8}\), P(B) = \(\frac{5}{8}\) and P(A ∪ B) = \(\frac{3}{4}\) then P(A/B). P(A’/B) is equal to
(a) \(\frac{6}{25}\)
(b) \(\frac{3}{20}\)
(c) \(\frac{3}{8}\)
(d) \(\frac{2}{5}\)
Answer:
(a) \(\frac{6}{25}\)

We know that
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

Question 15.
IfA and B are independent events, then P (A ∩ B) is equal to
(a) P(A) + P(B)
(b)P(A) – P(B)
(c) P(A).P(B)
(d) P(B)
Answer:
(c) P(A).P(B)

Given A and B are independent events
P (A ∩ B) = P (A) P (B)

Question 16.
If A and B are two independent events with P (A) = \(\frac{3}{5}\) and P (B) = \(\frac{4}{9}\), then P (A’ ∩ B’) is equal to
(a) \(\frac{4}{15}\)
(b) \(\frac{8}{15}\)
(c) \(\frac{1}{3}\)
(d) \(\frac{2}{9}\)
Answer:
(d) \(\frac{2}{9}\)

Since A and B are independent events then so are A’ and B’
∴ P (A’ ∩ B’) = P (A’) . P (B’)
= [1 – P(A)][1 – P(B)]
= [1 – \(\frac{3}{5}\)][1 – \(\frac{4}{9}\)]
= \(\frac{2}{5} \times \frac{5}{9}=\frac{2}{9}\)

Question 17.
If A and fi are two independent events such that P (A ∪ B’) = 0.8 and P (A) = 0.3 then P (B) is equal to
(a) \(\frac{2}{3}\)
(b) \(\frac{2}{7}\)
(c) \(\frac{1}{8}\)
(d) \(\frac{3}{8}\)
Answer:
(b) \(\frac{2}{7}\)

Given A and B are independent events then so are A and B’.
∴ P(A ∪ B’) = P (A) + P (B’) – P(A ∩ B’)
0.8 = 0.3 + P (B’) – P (A) P (B’)
⇒ 0.5 = P(B’) [1 – 0.3]
⇒ P(B’) = \(\frac{0.5}{0.7}=\frac{5}{7}\)
⇒ 1 – P(B) = \(\frac{5}{7}\)
⇒ P(B) = 1 – \(\frac{5}{7}=\frac{2}{7}\)

Question 18.
If A and B are two events such that P(A) = \(\frac{1}{4}\) P(A/B) = \(\frac{1}{2}\) and P(B/A) = \(\frac{2}{3}\) then P (B) is equal to
(a) \(\frac{1}{3}\)
(b) \(\frac{1}{6}\)
(c) \(\frac{1}{2}\)
(d) \(\frac{2}{3}\)
Answer:
(a) \(\frac{1}{3}\)

Question 19.
If S is the sample space of a random experiment, S = A ∪ B and P(A) = \(\frac{1}{3}\) P(B), where A and B are mutually exclusive events, then P (A) is equal to
(a) \(\frac{1}{2}\)
(b) \(\frac{3}{4}\)
(c) \(\frac{1}{4}\)
(d) \(\frac{3}{8}\)
Answer:
(c) \(\frac{1}{4}\)

Given S = A ∪ B
⇒ P(S) = 1
⇒ P (A) + P (B) = 1
[∵ A and B are mutually exclusive events A ∩ B = Φ ⇒ P(A ∩ B) = 0]
⇒ \(\frac{1}{3}\)P(B) + P(B) = 1
⇒ \(\frac{4}{3}\)P(B) = 1
⇒ P(B) = \(\frac{3}{4}\)
P(A) = 1 – P(B) = 1 – \(\frac{3}{4}\) = \(\frac{1}{4}\)

Question 20.
If two events are independent, then
(a) they must be mutually exclusive
(b) sum of their probability must be 1
(c) (a) and (b) are both correct
(d) none of the above is correct
Answer:
(d) none of the above is correct

Given two events are independent.
Then P (A ∩ B) = P (A) . P (B) and A, B are mutually exclusive then
P(A ∩ B) = 0

Question 21.
If A and B are two events such that
P(\(\overline{(A \cup B)}\)) = \(\frac{1}{6}\), P (A ∩ B) = \(\frac{1}{4}\) and P(Ā) = \(\frac{1}{4}\) then events A and B are 4
(a) independent but not equally likely
(b) mutually exclusive and independent
(c) equally likely and mutually exclusive
(d) equally likely but not independent
Answer:
(a) independent but not equally likely

Given P(\(\overline{(A \cup B)}\)) = \(\frac{1}{6}\)
P(A ∪ B) = 1 – P(A ∪ B) = 1 – 7 = 7
and P (A) = 1 – P (Ā) = 1 – \(\frac{1}{4}=\frac{3}{4}\)
We know that,
P (A ∪ B) = P (A) + P (B) – P (A ∩ B)
⇒ \(\frac{5}{6}=\frac{3}{4}\) + P(B) – \(\frac{1}{4}\)
⇒ P(B) = \(\frac{5}{6}-\frac{1}{2}=\frac{2}{6}=\frac{1}{3}\)
Here P(A ∩ B) = \(\frac{1}{4}\) ≠ 0
∴ A and B are not mutually exclusive.
Now P (A) P (B) = \(\frac{3}{4} \times \frac{1}{3}=\frac{1}{4}\) = P (A ∩ B)
Thus A and B are independent events. Since P (A) ≠ P (B) so A and B are not equally likely.

Question 22.
If P(A) = \(\frac{1}{3}\), P(B) = \(\frac{1}{2}\) and P(A ∪ B) = \(\frac{5}{6}\) then events A and B are
(a) independent
(b) independent and mutually exclusive
(c) mutually exclusive
(d) none of these
Answer:
(c) mutually exclusive

We know that
P (A ∪ B) = P (A) + P (B) – P (A ∩ B)
⇒ \(\frac{5}{6}=\frac{1}{3}+\frac{1}{2}\) – P(A ∩ B)
⇒ P(A ∩ B) = \(\frac{5}{6}-\frac{5}{6}\) = 0
∴ A and B are mutually exclusive events.

Question 23.
If P (A ∩ B) = \(\frac{1}{3}\) P (A ∪ B) = \(\frac{5}{6}\) and P (A) = \(\frac{1}{2}\) then which one of the following is correct ?
(a) A and B are independent events
(b) A and B are mutually exclusive events
(c) A and B are equally likely events
(d) none of these
Answer:
(a) A and B are independent events

Given P(A) = \(\frac{1}{3}\); P(A B) = \(\frac{5}{6}\) and P (A) = \(\frac{1}{2}\)
We know that
P (A ∪ B) = P (A) + P (B) – P (A ∩ B)
⇒ \(\frac{5}{6}=\frac{1}{2}\) + P(B) – \(\frac{1}{3}\)
⇒ P(B) = \(\frac{5}{6}+\frac{1}{3}-\frac{1}{2}=\frac{5+2-3}{6}=\frac{2}{3}\)
Now P(A).P(B) = \(\frac{1}{2} \times \frac{2}{3}=\frac{1}{3}\)
= P(A ∩ B)
Thus A and B are independent.

Question 24.
An experiment yields 3 mutually exclusive and exhaustive events A, B and C. If P (A) = 2 P (B) = 3 P (C), then P (A) is equal to
(a) \(\frac{1}{11}\)
(b) \(\frac{2}{11}\)
(c) \(\frac{3}{11}\)
(d) \(\frac{6}{11}\)
Answer:
(d) \(\frac{6}{11}\)

Since A, B and C are mutually exclusive and exhaustive events.
∴ P (A) + P (B) + P (C) = 1
⇒ P(A) + \(\frac{1}{2}\)p(A) + \(\frac{1}{3}\)P(A) = l
⇒P(A) \(\left[1+\frac{1}{2}+\frac{1}{3}\right]\) = 1
⇒ P (A) \(\left[\frac{6+3+2}{6}\right]\) = 1
⇒ P (A) = \(\frac{6}{11}\)

Question 25.
If P (A ∪ B) = 0.8 and P (A ∩ B) = 0.3, then P (A’) + P (B’) is equal to
(a) 0.7
(b) 0.9
(c) 0.5
(d) 1.1
Answer:
(b) 0.9

We know that,
P (A ∪ B) = P (A) + P (B) – P (A ∩ B)
⇒ 0.8 = P (A) + P (B) – 0.3
⇒ P (A) + P (B) =1.1

∴ P (A’) + P (B’) = 1 – P (A) + 1 – P (B)
= 2 – [P(A) + P(B)]
= 2 – 1.1
= 0.9

Question 26.
If A and B are two events such that P(A ∪ B) = \(\frac{3}{4}\) P (A ∩ B) = \(\frac{1}{4}\) and P(Ā) = \(\frac{2}{3}\), then P(Ā ∩ B) is equal to
(a) \(\frac{3}{8}\)
(b) \(\frac{5}{8}\)
(c) \(\frac{5}{12}\)
(d) \(\frac{1}{4}\)
Answer:
(c) \(\frac{5}{12}\)

Given P (A) = \(\frac{2}{3}\)
⇒ P(A) = 1 – P(A) = 1 – \(\frac{2}{3}=\frac{1}{3}\)

We know that,
P (A ∪ B) = P (A) + P (B) – P (A ∩ B)
⇒ \(\frac{3}{4}=\frac{1}{3}\) + P(B) – \(\frac{1}{4}\)
⇒ P (B) = \(\frac{3}{4}-\frac{1}{3}+\frac{1}{4}=1-\frac{1}{3}=\frac{2}{3}\)
Thus, P(A ∩ B) = P(B) – P(A ∩ B)
= \(\frac{2}{3}-\frac{1}{4}=\frac{5}{12}\)

Question 27.
Five persons A, B, C, D and E are standing in a queue of a ration shop. The probability that A and E are always together is
(a) \(\frac{1}{4}\)
(b) \(\frac{2}{3}\)
(c) \(\frac{2}{5}\)
(d) \(\frac{3}{5}\)
Answer:
(c) \(\frac{2}{5}\)

Total no. of ways in which five persons standing in a queue = 5! = 120
When A and E occur together it forms a group.
Now this group and remaining three persons can be arranged in 4! ways.
Now (AE) group can be arranged in 2! ways.
∴ Total No. of favourable outcomes = 4! × 2!
∴ required prob. = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\)
= \(\frac{4 ! \times 2 !}{5 !}=\frac{2}{5}\)

Question 28.
5 boys and 5 girls are sitting in a row randomly. The probability that boys and girls sit alternately, is
(a) \(\frac{5}{126}\)
(b) \(\frac{4}{126}\)
(c) \(\frac{3}{126}\)
(d) \(\frac{1}{126}\)
Answer:
Total no. of ways in which 5 boys and 5 girls are sitting in row =10!
Now 5 boys and 5 girls are sitting alternatively so they can sit in following schemes :
G₁B₁ G₂B₂ G₃B₃ G₄B₄ G₅B₅
B₁G₁ B₂G₂ B₃G₃ B₄G₄ B₅G₅
Thus total no. of favourable outcomes
= 2 × 5! × 5!
[since 5 boys can be arranged in 5! ways and we have to arrange 5 girls in 5 places in 5! ways]
∴ required prob. = \(\frac{2 \times 5 ! \times 5 !}{10 !}\)
= \(\frac{2 \times 5 \times 4 \times 3 \times 2}{10 \times 9 \times 8 \times 7 \times 6}=\frac{1}{126}\)

Question 29.
Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house, is
(a) \(\frac{7}{9}\)
(b) \(\frac{8}{9}\)
(c) \(\frac{1}{9}\)
(d) \(\frac{2}{9}\)
Answer:
(c) \(\frac{1}{9}\)

Each person can be apply for any of three houses and this can be done in 3 ways.
Thus, total no. of ways in which three persons applies for three houses without consulting others = 3 × 3 × 3 = 27
Now all the three persons can applies for H₁, H₂, H₃ in 3 ways.
Therefore, reqd. prob. = \(\frac{3}{27}\) = \(\frac{1}{9}\)

Question 30.
Out of 40 consecutive natural numbers, two numbers are chosen at random. Probability that the sum of numbers is odd, is
(a) \(\frac{14}{39}\)
(b) \(\frac{20}{39}\)
(c) \(\frac{1}{2}\)
(d) none of these
Answer:
(b) \(\frac{20}{39}\)

The total no. of ways of selecting 2 natural num bexs outof 40 = ⁴⁰C₂
Since we know that sum of odd and even numbers is odd.
Now there are 20 odd and 20 even numbers.
Thus total no. of favourable outcomes = ²⁰C₁ x ²⁰C₁
∴ required probability = \(\frac{{ }^{20} \mathrm{C}_1 \times{ }^{20} \mathrm{C}_1}{{ }^{40} \mathrm{C}_2}\)
= \(\frac{20 \times 20}{\frac{40 \times 39}{2}}=\frac{20}{39}\)

Question 31.
A bag contains 8 red and 7 black balls. Two balls are drawn at random. The probability that both the balls are of the same colour, is
(a) \(\frac{14}{15}\)
(b) \(\frac{11}{15}\)
(c) \(\frac{7}{15}\)
(d) \(\frac{4}{15}\)
Answer:
Total no. of ways of drawing 2 balls out of 15 = ¹⁵C₂
[∵ Total no. of balls = 8 + 7 = 15]
E : event that both balls are of same colour i.e. both are red or both are black
∴ n(E) = ⁸C₂ + ⁷C₂

Question 32.
A bag contains 5 red and 3 blue balls. If three balls are drawn at random without replacement, the probability of getting exactly one red ball is
(a) \(\frac{45}{196}\)
(b) \(\frac{135}{392}\)
(c) \(\frac{15}{56}\)
(d) \(\frac{15}{29}\)
Answer:
(c) \(\frac{15}{56}\)

No. of red balls = 5
No. of blue balls = 3
Total no. of balls = 5 + 3 = 8
Thus required prob.
= P (R₁B₂B₃) + P (B₁R₂B₃) + P (B₁B₂R₃)

Question 33.
A box contains 3 orange balls, 3 green balls and 2 blue balls. Three balls are drawn at random from the box without replacement The probability of drawing 2 green balls and one blue ball is
(a) \(\frac{3}{28}\)
(b) \(\frac{2}{21}\)
(c) \(\frac{1}{28}\)
(d) none of these
Answer:
(a) \(\frac{3}{28}\)

No. of orange balls = 3
No. of green balls = 3
No. of blue balls = 2
∴ Total no. of balls = 3 + 3 = 2 = 8
Thus required prob. of drawing 2 green balls and one blue ball
= \(\frac{{ }^3 \mathrm{C}_2 \times{ }^2 \mathrm{C}_1}{{ }^8 \mathrm{C}_3}=\frac{3 \times 2 \times 6}{8 \times 7 \times 6}=\frac{3}{28}\)

Question 34.
A bag contains 5 red and 3 blue balls. If three balls are drawn one by one without replacement from the bag, then the probability of drawing exactly two red balls of the three balls, the first ball being red, is
(a) \(\frac{1}{3}\)
(b) \(\frac{4}{7}\)
(c) \(\frac{5}{14}\)
(d) \(\frac{5}{28}\)
Answer:
(c) \(\frac{5}{14}\)

Total no. of balls = 5 + 3 = 8
∴ required probability
= P(R₁R₂B₃) + P(R₁B₂R₃)
= \(\frac{5}{8} \times \frac{4}{7} \times \frac{3}{6}+\frac{5}{8} \times \frac{3}{7} \times \frac{4}{6}\)
= \(\frac{120}{8 \times 7 \times 6}=\frac{20}{56}=\frac{5}{14}\)

Question 35.
Assume that in a family, each child is equally likely to be a boy or a girl. A family with three children is chosen at random. The probability that the eldest child is a girl given that the family has atleast one girl is
(a) \(\frac{1}{2}\)
(b) \(\frac{1}{3}\)
(c) \(\frac{2}{3}\)
(d) \(\frac{4}{7}\)
Answer:
(d) \(\frac{4}{7}\)

Given a family have three children
∴ S = {BBB, BBQ BGB, GBB, GGB, GBQ BGQ GGG}
⇒ n (S) = 8
Let A : event that eldest child is girl.
B : event that family has atleast one girl
∴ A = {GBB, GGB, GBG GGG}
B = {BBG BGB, GBB, GGB, GBQ BGQ GGG}
⇒ A ∩ B = {GBB, GGB, GBG GGG}

Question 36.
A die is thrown and a card is selected at random from a deck of 52 playing cards. The probability of getting an even number on the die and a spade card is
(a) \(\frac{1}{2}\)
(b) \(\frac{1}{4}\)
(c) \(\frac{1}{8}\)
(d) \(\frac{3}{13}\)
Answer:
(c) \(\frac{1}{8}\)

Let A : event of getting an even number = {2, 4, 6}
B : event that drawn crd is a spade card.
∴ P(A) = \(\frac{3}{6}=\frac{1}{2}\) and P(B) = \(\frac{1}{2} \times \frac{1}{4}=\frac{1}{8}\)
Thus required prob. = P (A) . P (B)
= \(\frac{1}{2} \times \frac{1}{4}=\frac{1}{8}\)

Question 37.
A and B play a game where each is asked to select a number from 1 to 25. If the two numbers match, both of them win a prize. The probability that they will not win a prize in a single trial, is
(a) \(\frac{1}{25\)
(b) \(\frac{24}{25}\)
(c) \(\frac{2}{25}\)
(d) none of these
Answer:
(b) \(\frac{24}{25}\)

It is given that, if A and B select a same number then both of them win prize.
Now both of them will not get a prize, if A choose any number a then B can choose any no other than a so there 24 such ways.
∴ required prob. = \(\frac{24}{25}\)

Question 38.
A bag contains 4 black balls and 3 white balls. Balls are picked up at random and placed in a row. The probability that the balls are alternately of different colours is
(a) \(\frac{1}{35}\)
(b) \(\frac{3}{35}\)
(c) \(\frac{2}{7}\)
(d) none of these
Answer:
Total no. of balls = 4 + 3 = 7
Total no. of ways in which all balls can be arranged = 7!
Now balls are placed alternatively of different Colours and this can be done in following ways :
B₁W₁B₂W₂B₃W₃B₄
The four black balls can be arranged in 4! ways and three white balls placed in three places in 3! ways.
No. of favourable outcomes = 3! × 4!
∴ required prob. = \(\frac{3 ! \times 4 !}{7 !}=\frac{6}{7 \times 6 \times 5}=\frac{1}{35}\)

Question 39.
Three numbers are chosen at random from 1 to 20. The probability that they are consecutive is
(a) \(\frac{1}{190}\)
(b) \(\frac{1}{120}\)
(c) \(\frac{3}{190}\)
(d) \(\frac{5}{190}\)
Answer:
The total no. of ways of selecting 3 numbers out of 20 = 20C3 = n (S)
E : event that the chosen numbers are three consecutive numbers
= {(123), (234), (343), ,(18 19 20)}
∴ n(E) = 18
Thus, required prob. = \(\frac{n(\mathrm{E})}{n(\mathrm{~S})}\)
= \(\frac{18}{{ }^{20} C_3}=\frac{18}{\frac{20 \times 19 \times 18}{6}}=\frac{3}{190}\)

Question 40.
A person draws out two balls successively from a bag containing 6 red and 4 white balls. The probability thatatleast one of them will be red is
(a) \(\frac{12}{90}\)
(b) \(\frac{30}{90}\)
(c) \(\frac{48}{90}\)
(d) \(\frac{78}{90}\)
Answer:
(d) \(\frac{78}{90}\)

Total no. of balls = 6 + 4 = 10
required prob. = P (RB) + P (BR) + P (RR)
= \(\frac{6}{10} \times \frac{4}{9}+\frac{4}{10} \times \frac{6}{9}+\frac{6}{10} \times \frac{5}{9}=\frac{78}{90}\)

Question 41.
A bag contains 5 brown and 4 black socks. Aman pulls out two socks. The probability that these are of the same colour is
(a) \(\frac{5}{108}\)
(b) \(\frac{18}{108}\)
(c) \(\frac{30}{108}\)
(d) \(\frac{48}{108}\)
Answer:
(d) \(\frac{48}{108}\)

Total no. of ways of selecting 2 socks out of 9 socks = ⁹C₂
Required prob. that the selected socks are of same colour = P (BB) + P (B₁B₂)

Question 42.
The probability that in the toss of two dice we obtain the sum 7 or 11 is
(a) \(\frac{1}{18}\)
(b) \(\frac{1}{6}\)
(c) \(\frac{2}{9}\)
(d) \(\frac{23}{108}\)
Answer:
(c) \(\frac{2}{9}\)

In a toss of two dice
total no. of outcomes = 6² = 36 = n (S)
E : event that sum is 7 or 11 = {(1,6), (2, 5), (3,4), (4,3), (5,2), (6,1), (5, 6), (6, 5)}
n (E) = 8
Thus required probability = \(\frac{n(\mathrm{E})}{n(\mathrm{~S})}\)
= \(\frac{8}{36}=\frac{2}{9}\)

Question 43.
A pair of dice is thrown. If 5 appears on atleast one of the dice, then the probability that the sum is 10 or greater is
(a) \(\frac{11}{36}\)
(b) \(\frac{2}{9}\)
(c) \(\frac{3}{11}\)
(d) \(\frac{1}{12}\)
Answer:
(c) \(\frac{3}{11}\)

When a pair of dice is thrown
∴ n (S) = 36
Let E : event that 5 appears on atleast one the dice
= {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (1,5), (2, 5), (3,5), (4, 5), (6, 5)}
and F = event that sum is 10 or greater = {(4,6), (6,4), (5, 5), (5, 6), (6, 5), (6, 6)}
E ∩ F = {(5, 5), (5. 6), (6, 5)}
n (E ∩ F) = 3 ; n (E) = 11 ; n (F) = 6
Thus required prob. = P (F/E)
= \(\frac{P(E \cap \mathrm{F})}{P(E)}=\frac{n(E \cap F)}{n(\mathrm{E})}=\frac{3}{11}\)

Question 44.
Two dice are thrown. It is known that the sum of numbers on the dice was less than 6, the probability that the sum is less than 3 is
(a) \(\frac{1}{18}\)
(b) \(\frac{5}{18}\)
(c) \(\frac{1}{10}\)
(d) \(\frac{2}{5}\)
Answer:
(c) \(\frac{1}{10}\)

When two dice are thrown.
Then n (S) = 6² = 36
Let E : event that the sum of numbers on
dice was < 6 = {(1,1), (1,2), (1,3), (1,4), (2, 1), (3, 1), (4, 1)}
F : event that sum is less than 3 = {(1, 1)}
E ∩ F = {(1, 1)}
⇒ n (E) = 7 ; n (F) = 1 = n (E ∩ F)
.’. required prob. = P (F/E)
\(\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{E})}=\frac{n(\mathrm{E} \cap \mathrm{F})}{n(\mathrm{E})}=\frac{1}{7}\)

Question 45.
In a college, 30% students fail in physics, 25% fail in mathematics and 10% fail in both subjects. One student is chosen at random. The probability that she fails in physics if she has failed in mathematics is
(a) \(\frac{1}{10}\)
(b) \(\frac{2}{5}\)
(c) \(\frac{9}{20}\)
(d) \(\frac{1}{3}\)
Answer:
(b) \(\frac{2}{5}\)

Let us define the events are as follows :
A : students fail in Physics
B : students fail in Maths
Then P (A) = \(\frac{30}{100}\) ; P (B) = \(\frac{25}{100}\)
P(A ∩ B) = \(\frac{10}{100}\)
required prob. = P (A/B)
= \(\frac{P(A \cap B)}{P(B)}=\frac{\frac{10}{100}}{\frac{25}{100}}=\frac{2}{5}\)

Question 46.
The probability of throwing 16 in one throw with three dice is
(a) \(\frac{1}{36}\)
(b) \(\frac{1}{18}\)
(c) \(\frac{1}{72}\)
(d) \(\frac{1}{9}\)
Answer:
(a) \(\frac{1}{36}\)

When three dice are thrown.
Then total no. of outcomes = n (S)
= 6³ = 216
Let E : event of throwing 16 in one throw of three dice
= {(4, 6, 6), (6,4, 6), (6, 6,4), (5, 5, 6), (5, 6, 5), (6, 5, 5)}
∴ n (E) = 6
Thus required probability = \(\frac{n(\mathrm{E})}{n(\mathrm{~S})}\)
= \(\frac{6}{216}=\frac{1}{36}\)

Question 47.
X speaks truth in 60% and Y in 50% of the cases. The probability that they contradict each other while narrating the same fact is
(a) \(\frac{1}{4}\)
(b) \(\frac{1}{3}\)
(c) \(\frac{1}{2}\)
(d) \(\frac{2}{3}\)
Answer:
(c) \(\frac{1}{2}\)

P (X speaks truth) = \(\frac{60}{100}\)
P (Y speaks truth) = \(\frac{50}{100}\)
P (X tell a lie) = 1 – \(\frac{60}{100}=\frac{40}{100}\)
P (Y tell a lie) = \(\frac{50}{100}\)
∴ required probability
= P (X speaks truth) P (Y tell a lie) + P (X tell a lie) P (Y speaks truth)
=\(\frac{60}{100} \times \frac{50}{100}+\frac{40}{100} \times \frac{50}{100}=\frac{5000}{10000}=\frac{1}{2}\)

Question 48.
If the probability for A to fail in an examination is 0.2 and that for B is 0.3, then the probability that either A fails or B fail is
(a) 0.5
(b) 0.06
(c) 0.56
(d) 0.44
Answer:
Given P (A fails) = 0.2 and P(B fails) = 0.3
∴ P (A fails or B fails) = P (A fails) + P(B fails) – P (A fails and B fails)
Since A and B are independent events = P (A fails) + P (B fails) – P (A fails) × P(B fails)
= 0.2 + 0.3 – 0.2 × 0.3
= 0.5 – 0.06
= 0.44

Question 49.
A problem ¡n mathematics ¡s given to three students A, B, C and their chances of solving the problem are \(\frac{1}{2}, \frac{1}{3}\) and \(\frac{1}{4}\) respectively. The probability that the problem is solved is
(a) \(\frac{3}{4}\)
(b) \(\frac{1}{2}\)
(c) \(\frac{2}{3}\)
(d) \(\frac{5}{8}\)
Answer:
Given P (A) = \(\frac{1}{2}\); P (B) = \(\frac{1}{3}\);
P(C) = \(\frac{1}{4}\)
P(A) = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\); P(B) = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\) and P(C) = 1 – \(\frac{1}{4}\) = \(\frac{3}{4}\)
Thus required prob. that problem is solved
= 1 – P (Ā B̄ C̄)
= 1 – P(Ā) P(B̄) P(C̄)
= 1 – \(\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4}\)
= 1 – \(\frac{1}{4}=\frac{3}{4}\)

Question 50.
Three persons A, B and C fire a target in turn. Their probabilities of hitting the target arc 0.4, 0.3 and 0.2 respectively. The probability of two hits is
(a) 0.024
(b) 0.188
(c) 0.336
(d) 0.452
Answer:
(b) 0.188

Given P (A hits a target) = P (A) = 0.4 ; P (B) = 0.3 ; P (C) = 0.2
∴ required prob. = P (ABC̄) + P (AB̄C) + P (ĀBC)
= P (A) P (B) P (C̄) + P (A) P (B̄) P (C) + P (Ā) P (B) P (C)
= (0.4) (0.3) (1 – 0.2) + (0.4) (1 – 0.3) (0.2) + (1 – 0.4) (0.3) (0.2)
= (0.4) (0.3) (0.8) + (0.4) (0.7) (0.2) + (0.6) (0.3) (0.2)
= 0.096 + 0.056 + 0.036
= 0.188

Question 51.
A and B toss 3 coins. The probability that they both obtain the same number of heads is
(a) \(\frac{1}{9}\)
(b) \(\frac{3}{16}\)
(c) \(\frac{5}{16}\)
(d) \(\frac{3}{8}\)
Answer:
(c) \(\frac{5}{16}\)

Let X be the number of heads obtained by A and Y be the no. of heads obtained by B
p = prob. of getting a head in a angle toss of a coin = \(\frac{1}{2}\)
q = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\) and n = 3
Then by Binomial distribution
P (X = r) = ⁿC_r p^r q^n-r
= ³C_r\(\left(\frac{1}{2}\right)^r\left(\frac{1}{2}\right)^{3-r}\)
= ³C_r\(\left(\frac{1}{2}\right)^3\)

Thus required probability that A and B obtain same no. of heads
= P (X = 0) P (Y = 0) + P (X = 1) P (Y = 1) + P (X = 2) P (Y = 2) + P (X = 3) P (Y = 3)

Question 52.
There are two bags. One bag contains 4 w hite balls and 2 black balls. The other bag contains 7 white and 3 black balls. If a bag is selected at random and from it, a ball is drawn, then the probability that the ball is black, is
(a) \(\frac{1}{3}\)
(b) \(\frac{1}{5}\)
(c) \(\frac{3}{20}\)
(d) \(\frac{7}{20}\)
Answer:
Bag-I contains 3 white and 2 black balls, Bag-II contains 7 white and 3 black balls
Let E₁ : bag-I is chosen;
E₂ : bag-11 is chosen;
E : chosen ball is black
Then P (E₁) = P (E₂) = \(\frac{1}{2}\);
P (E/E₁) = \(\frac{2}{5}\);
P (E/E₂) = \(\frac{3}{10}\)
∴ required prob. = P (E)
= P(E₁). P (E/E₁) + P (E₂) . P (E/E₂)
= \(\frac{1}{2} \times \frac{2}{5}+\frac{1}{2} \times \frac{3}{10}=\frac{4+3}{20}=\frac{7}{20}\)

Question 53.
From the set {1,2,3,4,5}, two numbers a and b(a ≠ b) are chosen at random. The probability that \(\frac{a}{b}\) is an integer is
(a) \(\frac{1}{3}\)
(b) \(\frac{1}{4}\)
(c) \(\frac{1}{2}\)
(d) \(\frac{3}{5}\)
Answer:
(b) \(\frac{1}{4}\)

Given set (1,2,3,4,5)
∴ S = {(1, 2), (1, 3), (I, 4), (1, 5), (2, 1), (2, 3), (2, 4), (2, 5), (3, 1), (3, 2), (3, 4), (3, 5), (4, 1), (4, 2), (4, 3), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4))
n (S) = 20
Now two numbers a and b (a ≠ b) are chosen s.t \(\frac{a}{b}\) is an integer.
∴ No. of favourable outcomes = 5
{(2, 1),(3, 0,(4, 0,(5, 0,(4, 2)}
required probability = \(\frac{5}{20}=\frac{1}{4}\)

Question 54.
A card is picked at random from a pack of 52 playing cards. Given that the picked card is a queen, the probability of this card to be a card of hearts is
(a) \(\frac{1}{3}\)
(b) \(\frac{4}{13}\)
(c) \(\frac{1}{4}\)
(d) \(\frac{1}{2}\)
Answer:
(c) \(\frac{1}{4}\)

Let A : event that picked card is green
B : event that chosen card is a card of hearts
P(A) = \(\frac{4}{52}=\frac{1}{13}\)

P (A n B) = prob. of getting queen of heart
= \(\frac{1}{52}\)

required prob. = P (B/A) = \(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}\)
= \(\frac{\frac{1}{52}}{\frac{1}{13}}=\frac{1}{4}\)

Question 55.
If A and B are two independent events with P (A) = \(\frac{1}{3}\) and P(B) = \(\frac{1}{4}\) then P (B’/A) is equal to
(a) \(\frac{1}{4}\)
(b) \(\frac{1}{3}\)
(c) \(\frac{3}{4}\)
(d) 1
Answer:
(c) \(\frac{3}{4}\)

Given A and B are independent event then so are B’ and A.
∴ P(B’/A) = \(\frac{P\left(B^{\prime} \cap A\right)}{P(A)}\)
= \(\frac{P\left(B^{\prime}\right) P(A)}{P(A)}\) = P(B’)
= 1 – P(B)
= 1 – \(\frac{1}{4}=\frac{3}{4}\)

Question 56.
The probability distribution of a discrete random variable X is given below :

Then, E (X) is equal to
(a) 6
(b) 4
(c) 3
(d) -5
Answer:
(b) 4

∴ E (X) = ΣX P(X) = 30 × \(\frac{1}{5}\) + 10 × \(\frac{1}{5}\) + 10 × \(\frac{1}{2}\)
= 6 + 3 – 5
= 4

Question 57.
For the probability distribution:

E(X) is equal to
(a) 3
(b) 5
(c) 7
(d) 10
Answer:
(d) 10

∴ E(X²) = ΣX² P(X)
= -4 × 0.1 – 3 × 0.2 – 2 × 0.3 – 1 × 0.2 + 0 × 0.2
= -0.4 – 0.6 – 0.6 – 0.2
= -1.8

Question 58.
For the following probability distribution:

E(X²) is equal to
(a) 3
(b) 5
(c) 7
(d) 10
Answer:
(d) 10

∴ E(X²) = ΣX² P(X)
= 1² × \(\frac{1}{10}\) + 2² × \(\frac{1}{5}\) + 3² × \(\frac{3}{10}\) + 4² × \(\frac{2}{5}\)
= \(\frac{1}{10}+\frac{4}{5}+\frac{27}{10}+\frac{32}{5}\)
= \(\frac{1+8+27+64}{10}=\frac{100}{10}\)
= 10

Question 59.
For a random variable X, E(X²) = 11. Then variance of X is
(a) 8
(b) 5
(c) 2
(d) 1
Answer:
(c) 2

Given E(X) = 3 and E(X²) = 11
∴ var (X) = E(X²) – (E(X))²
= 11 – 3²
= 2

Question 60.
A random variable ‘X’ has the following probability distribution :

The value of k is
(a) -1
(b) –\(\frac{1}{10}\)
(c) 1
(d) \(\frac{1}{10}\)
Answer:
(d) \(\frac{1}{10}\)

Since ΣP (X) = 1
⇒ P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6) + P (X = 7) = 1
⇒ k + k + 2k + 3k + k² + 2k² + 7k² + 2k = 1
⇒ 10k + 9k – 1 = 0
⇒ (k + 1) (10k – 1) = 0
⇒ k = -1, \(\frac{1}{10}\)
but k > 0 otherwise P (X = 0) < 0
k = \(\frac{1}{10}\)

Question 61.
A dice is rolled thrice. If the event of getting an even number is a success, then the probability of getting atleast two successes is
(a) \(\frac{7}{8}\)
(b) \(\frac{1}{4}\)
(c) \(\frac{2}{3}\)
(d) \(\frac{1}{2}\)
Answer:
(d) \(\frac{1}{2}\)

p = prob. of getting an even number in single throw of dice = \(\frac{3}{6}\) = \(\frac{1}{2}\)
q = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\) ;n = 3.
Since the dice is rolled thrice so events are independent so it is a problem of binomial distribution.
P (X = r) = ⁿC_rp^rq^n-r
= ³C_r \(\left(\frac{1}{2}\right)^r\left(\frac{1}{2}\right)^{3-r}\)
= ³C_r \(\left(\frac{1}{2}\right)^3\)
Thus required prob. = P(X ≥ 2)
= P(X = 2) + P(X = 3)
= ³C₂\(\left(\frac{1}{2}\right)^3\) + ³C₃\(\left(\frac{1}{2}\right)^3\)
= (3 + 1)\(\frac{1}{8}=\frac{1}{2}\)

Question 62.
A dice is thrown twice, the probability of occurring of 5 atleast once is
(a) \(\frac{11}{36}\)
(b) \(\frac{7}{12}\)
(c) \(\frac{35}{36}\)
(d) none of these
Answer:
(a) \(\frac{11}{36}\)

Let p = prob. of getting 5 in a single throw of dice
q = 1 – p = 1 – \(
Then by binomial distribution, we have
p (X = r) = ⁿC_rp^rq^n-r

Question 63.
A coin is tossed 10 times. The probability of getting exactly six heads is
(a) ¹⁰C₆
(b) [latex]\frac{511}{512}\)
(c) \(\frac{105}{512}\)
(d) \(\frac{21}{512}\)
Answer:
(c) \(\frac{105}{512}\)

Let p = prob. of getting head = \(\frac{1}{2}\)
q = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\); n = 10
since events are independent
So it is a problem of binomial distribution
with n = 10 ; P = q = \(\frac{1}{2}\)
Thus P (X = r) = ⁿC_rp^rq^n-r
= ¹⁰C_r\(\left(\frac{1}{2}\right)^r\left(\frac{1}{2}\right)^{10-r}\)
= ¹⁰C_r\(\left(\frac{1}{2}\right)^{10}\)
required prob. of getting exactly six heads
= P(X = 6) = ¹⁰C₆\(\left(\frac{1}{2}\right)^{10}\)
= \(\frac{10 \times 9 \times 8 \times 7}{24} \times \frac{1}{2^{10}}\)
= \(\frac{210}{1024}=\frac{105}{512}\)

Question 64.
A coin is tossed 4 times. The probability of getting atleast one head is
(a) \(\frac{1}{16}\)
(b) \(\frac{1}{8}\)
(c) \(\frac{7}{8}\)
(d) \(\frac{15}{16}\)
Answer:
(d) \(\frac{15}{16}\)

Let p = prob. of getting a head in a single toss = \(\frac{1}{2}\)
q = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\) and n = 4
Thus by binomial distribution, we have
P(X = r) = ⁿC_rp^rq^n-r
= ⁴C_r\(\left(\frac{1}{2}\right)^r\left(\frac{1}{2}\right)^{4-r}\)
= ⁴C_r\(\left(\frac{1}{2}\right)^4\)
required prob. of getting atleast one head
= P (X ≥ 1) = 1 – P (X = 0)
= 1 – ⁴C₀\(\left(\frac{1}{2}\right)^4\)
= 1 – \(\frac{1}{16}=\frac{15}{16}\)

Question 65.
Two dice are thrown n times in succession. The probability of obtaining a double six atleast once is
(a) \(\left(\frac{1}{36}\right)^n\)
(b) 1 – \(\left(\frac{35}{36}\right)^n\)
(c) \(\left(\frac{1}{12}\right)^n\)
(d) none of these
Answer:
(b) 1 – \(\left(\frac{35}{36}\right)^n\)

When two dice are thrown
Then total no. of outcomes = 6² = 36
Let p = prob. of getting a double six in single throw of two dice = \(\frac{1}{36}\)
q = 1 – p = 1 – \(\frac{1}{36}=\frac{35}{36}\)
Thus by binomial distribution, we have
P (X = r) = ⁿC_rp^rq^n-r
= ⁿC_r\(\)
required prob. = P (X ≥ 1) = 1 – P (X = 0)
=1 – ⁿC₀\(\left(\frac{1}{36}\right)^0\left(\frac{35}{36}\right)^n\)
=1 – \(\left(\frac{35}{36}\right)^n\)

Question 66.
A bag contains 2 white and 4 black balls. A ball is drawn 5 times with replacement. The probability that atleast 4 of the balls drawn are white is
(a) \(\frac{8}{243}\)
(b) \(\frac{10}{243}\)
(c) \(\frac{11}{243}\)
(d) \(\frac{32}{243}\)
Answer:
(c) \(\frac{11}{243}\)

Total no. of balls = 2 + 4 = 6
∴ P(W) = \(\frac{2}{2+4}=\frac{1}{3}\)
P(B) = \(\frac{4}{2+4}=\frac{2}{3}\)
Thus required probability = P (WWWWB) + P (WWWBW) + P (WWBWW) + P (WBWWW) + P (BWWWW) + P (WWWWW)
= 5 × \(\frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} \times \frac{2}{3}+\left(\frac{1}{3}\right)^5\)
= \(\frac{10+1}{3^5}=\frac{11}{243}\)

Question 67.
If in a trial the probability of a success is twice the probability of failure, the probability of atleast four successes in six trials is
(a) \(\frac{496}{729}\)
(b) \(\frac{400}{729}\)
(c) \(\frac{500}{729}\)
(d) \(\frac{600}{729}\)
Answer:
(a) \(\frac{496}{729}\)

Thus by binomial distribution, we have
P (X = r) = ⁿC_r p^r q^{n – r}
required prob. = P (X ≥ 4) = P (X = 4) + P (X = 5) + P (X = 6)

Question 68.
A box contains 20 identical balls of which 10 balls are white and 10 balls are red. The balls are drawn at random from the box one at a time with replacement. The probability that a white ball is drawn for the 4th time on the 7th draw is
(a) \(\frac{5}{64}\)
(b) \(\frac{5}{32}\)
(c) \(\frac{27}{32}\)
(d) \(\frac{1}{2}\)
Answer:
(b) \(\frac{5}{32}\)

Total no. of identical balls = 20
prob. of drawing a white ball = P (W) = \(\frac{10}{20}=\frac{1}{2}\) = P and P(B) = \(\frac{10}{20}=\frac{1}{2}\) = q
∴ prob. of drawing a white ball in 7th draw = \(\frac{1}{2}\)
Thus prob. of drawing a white ball for the 4th time in %h draw = prob. of drawing 3 white balls in 6 draws x prob. of drawing white ball in 7th draw

Question 69.
In a series of three trials, the probability of two successes in 9 times the probability of three successes. Then, the probability of success in each trial is
(a) \(\frac{3}{4}\)
(b) \(\frac{1}{2}\)
(c) \(\frac{1}{3}\)
(d) \(\frac{1}{4}\)
Answer:
Let p = probability of success
q = prob. of failure ; «= 3
Thus by binomial distribution,
P (X = r) = ⁿC_r p^r q^{n – r}
= ³C_r p^r q^{3 – r}
Given P (X = 2) = 9 P (X = 3)
³C₂ p² q = 9 x ³C₃ p³
⇒ 3p²q = 9p³
⇒ q = 3p
⇒ 1 -p = 3p

Let p = probability of success and q = prob . of failure
Given p = 2q and p + q = 1
⇒ 2q + q = 1
⇒ 2q = 1
⇒ q = \(\frac{1}{3}\)
p = \(\frac{2}{3}\) and n = 6
⇒ 4p = 1
⇒ p = \(\frac{1}{4}\)

Question 70.
A coin is tossed n times. The probability of getting head atleast once is 0.8, then the least value of n is
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(b) 3

Let p = prob. of getting head in a single toss = \(\frac{1}{2}\)
q = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
Thus by binomial distribution, we have
P(X = r) = ⁿC_r p^r q^{n – r}
= ⁿC_r \(\left(\frac{1}{2}\right)^r\left(\frac{1}{2}\right)^{n-r}\)
= ⁿC_r\(\left(\frac{1}{2}\right)^n\)

Given P(X ≥ 1) = 0.8
⇒ 1 – P(X = 0) = 0.8
⇒ 0.2 = P(X = 0)
⇒ⁿC₀\(\left(\frac{1}{2}\right)^n=\frac{1}{2^n}\)
⇒ 2ⁿ = \(\frac{1}{0.2}\) = 5
Clearly the least value of n = 3

Question 71.
A box contains 100 bulbs of which 10 are defective. The probability that out of a sample of 5 bulbs drawn one by one with replacement none is defective is
(a) \(\left(\frac{1}{2}\right)^5\)
(b) \(\frac{9}{10}\)
(c) \(\left(\frac{9}{10}\right)^5\)
(d) \(\left(\frac{1}{10}\right)^5\)
Answer:
(c) \(\left(\frac{9}{10}\right)^5\)

Let p = probability of getting defective bulb
= \(\frac{10}{100}=\frac{1}{10}\)
q = 1 – p = 1 – \(\frac{1}{10}=\frac{9}{10}\)

Here bulbs are drawn one by one with replacement so events are independent so it is a problem of binomial distribution with n = 5
Then by binomial distribution,
P(X = r) = ⁿC_r p^r q^{n – r}

Question 72.
A fair coin is tossed n times. If the probability of getting seven heads is equal to the probability of getting 9 heads, then the probability of getting two heads is
(a) \(\frac{2}{15}\)
(b) \(\frac{15}{2^8}\)
(c) \(\frac{15}{2^{13}}\)
(d) none of these
Answer:
(c) \(\frac{15}{2^{13}}\)

Let p = probability of getting head = \(\frac{1}{2}\)
P(X = r) = ⁿC_r p^r q^{n – r}

Question 73.
A fair coin is tossed 100 times. The probability of getting head an odd number of times is
(a) \(\frac{1}{2}\)
(b) \(\frac{1}{3}\)
(c) \(\frac{1}{8}\)
(d) \(\frac{3}{8}\)
Answer:
(a) \(\frac{1}{2}\)

Let p = prob. of getting head = \(\frac{1}{2}\)
q = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\); n = 100
Then by binomial distribution, we have
P(X = r) = ⁿC_r p^r q^{n – r}

Question 74.
For a binomial variate X, if n = 4 and P (X = 0) = \(\frac{16}{81}\) then P (X = 4) is
(a) \(\frac{1}{3}\)
(b) \(\frac{1}{27}\)
(c) \(\frac{1}{81}\)
(d) \(\frac{1}{16}\)
Answer:
(c) \(\frac{1}{81}\)

Question 75.
For a binomial variate X, if n = 3 and P (X = 1) = 12 P (X = 3), then the value of p is
(a) \(\frac{1}{2}\)
(b) \(\frac{1}{3}\)
(c) \(\frac{2}{3}\)
(d) \(\frac{2}{5}\)
Answer:
Given P (X = 1) = 12 P (X = 3); n = 3
We know that,
P (X = r) = ⁿC_r p^r q^{n – r} = ³C_r p^r q^{3 – r}
⇒ ³C₁ p q² = 12 x ³C₃ p³ q⁰
⇒ 3 pq² = 12p³
⇒ q² = 4p²
⇒ (1 – p)² = 4p²
⇒ 3p² + 2/7 – 1 = 0
⇒ (p + 1) (3p- 1) = 0
⇒ P = -1, \(\frac{1}{3}\)
Since p > 0
p = \(\frac{1}{3}\)

Question 76.
The probability of guessing atleast 8 correct answers out of 10 true/false questions
(a) \(\frac{7}{64}\)
(b) \(\frac{7}{128}\)
(c) \(\frac{7}{256}\)
(d) \(\frac{35}{1024}\)
Answer:
(b) \(\frac{7}{128}\)

p = prob. of getting answer to true/false question
= \(\frac{1}{2}\)
q = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\); n = 10

Thus by binomial distribution,
P (X = r) = ⁿC_r p^r q^{n – r}

Question 77.
A dice is thrown 100 times. If getting an even number is considered a success, then the variance of the number of successes is
(a) 10
(b) 20
(c) 25
(d) 50
Answer:
(c) 25

p = prob. of getting an even number
= \(\frac{3}{6}=\frac{1}{2}\)
q = 1 – p = 1 – \(\frac{1}{2}=\frac{1}{2}\) and n = 10
∴ Variance = npq = 100 x \(\frac{1}{2} \times \frac{1}{2}\) = 25

Question 78.
If the mean and standard deviation of a binomial distribution are 12 and 2 respectively, then the value of its parameter p is
(a) \(\frac{1}{4}\)
(b) \(\frac{1}{3}\)
(c) \(\frac{1}{2}\)
(d) \(\frac{2}{3}\)
Answer:
Mean = np = 12 ……….(1)
and S.D = \(\sqrt{n p q}\) = 2
npq = 4 …..(2)
From (1) and (2); we have
12q = 4
q = \(\frac{1}{3}\)
p = 1 – q = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)

Question 79.
If the mean and the variance of a probability distribution are 4 and 2 respectively, then the probability of two successes is
(a) \(\frac{1}{2}\)
(b) \(\frac{7}{64}\)
(c) \(\frac{219}{256}\)
(d) \(\frac{37}{256}\)
Answer:
(b) \(\frac{7}{64}\)

Given mean = np = 4 ….(1)
and npq = 2 …..(2)
From (1) and (2); we have
4q = 2
q = \(\frac{1}{2}\)
p = 1 – q = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
Thus from (1); n = 4 × 2 = 8
Required prob. = P (X = 2)
= ⁸C₂\(\)
[∵P(X = r) = ⁿC_r p^r q^{n – r}]
= \(\frac{8 \times 7}{2} \times \frac{1}{2^8}=\frac{7}{64}\)

Question 80.
If the mean and the variance of a bino¬mial distribution are 4 and 3 respectively, then the probability of six successes is
(a) ¹⁶C₆\(\left(\frac{1}{4}\right)^6\left(\frac{3}{4}\right)^{10}\)
(b) ¹⁶C₆\(\left(\frac{1}{4}\right)^{10}\left(\frac{3}{4}\right)^6\)
(c) ¹²C₆\(\left(\frac{1}{4}\right)^6\left(\frac{3}{4}\right)^6\)
(d) ¹²C₆\(\left(\frac{1}{4}\right)^6\left(\frac{3}{4}\right)^{10}\)
Answer:
(a) ¹⁶C₆\(\left(\frac{1}{4}\right)^6\left(\frac{3}{4}\right)^{10}\)

Given mean = np = 2
variance = npq = 1
From (1) and (2); we have
4q = 3
q = \(\frac{3}{4}\)
p = 1 – q = 1 – \(\frac{3}{4}=\frac{1}{4}\)
from (1); « = 16
Thus by binomial distribution, we have
P(X = r) = ⁿC_r p^r q^{n – r}
= ¹⁶C_r\(\left(\frac{1}{4}\right)^r\left(\frac{3}{4}\right)^{16-r}\) ……(1)

required prob. = P(X = 6)
= ¹⁶C₆\(\left(\frac{1}{4}\right)^6\left(\frac{3}{4}\right)^{10}\)

Question 81.
If the mean and the variance of a binomial distribution are 2 and 1 respectively, then the probability of atleast one success is
(a) \(\frac{2}{3}\)
(b) \(\frac{3}{4}\)
(c) \(\frac{7}{8}\)
(d) \(\frac{15}{16}\)
Answer:
(d) \(\frac{15}{16}\)

Given mean = np = 2
variance = npq = 1
From (1) and (2); we have
2q = 1
q = \(\frac{1}{2}\)

Question 82.
In a binomial distribution, the probability of getting a success is \(\frac{1}{4}\) and the standard deviation is 3. Then its mean is
(a) 6
(b) 8
(c) 12
(d) 48
Answer:
(c) 12

p = prob. of success = \(\frac{1}{4}\)