Interactive ISC Mathematics Class 12 Solutions Chapter 10 Probability Ex 10.6 engage students in active learning and exploration.

## ML Aggarwal Class 12 Maths Solutions Section A Chapter 10 Probability Ex 10.6

Question 1.

A random variable X has the following probability distribution :

(i) Determine the value of a.

(ii) Find P (X < 3), P (X ≥ 4), P (0 < X < 5).

Answer:

Since the random variable X has probability distribution

(i) Σp_{i} = 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

⇒ a + 4a + 3a + la + 8a + 10a + 6a + 9a – 1

⇒ 48a = 1

⇒ a = \(\frac{1}{48}\)

(ii) P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

= a + 4a + 3a = 8a

= \(\frac{8}{48}=\frac{1}{6}\)

P (X ≥ 4) = P (X = 4) + P (X = 5) + P (X = 6) + P (X = 7)

= 8a + 10a + 6a + 9a = 33a

= \(\frac{33}{48}=\frac{11}{16}\)

and P (0 < X < 5) = P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)

= 4a + 3a + 7a + 8a = 22 a

= \(\frac{22}{48}=\frac{11}{24}\)

Question 2.

A random variable X has the probability distribution :

Answer:

E = {X is a prime number} = {2, 3, 5, 7}

F = {X < 4} = {1, 2, 3}

∴ E ∪ F = {1, 2, 3, 5, 7}

Thus P(E ∪ F) = P(1) + P(2) + P (3) + P (5) + P (7)

= 0.15 + 0.23 + 0.12 + 0.2 + 0.07

= 0.77

Question 3.

A random variable X has the following probability distribution where k is some number :

(i) Determine the value of k.

(ii) Find P (X <2), P (X ≤ 2), P (X ≥ 2). (NCERT)

Answer:

We know that otherwise Σp_{i} = 1

⇒ k + 2k + 3k + 0 = 1

⇒ 6k = 1

⇒ k = \(\frac{1}{6}\)

(ii) P (X < 2) = P (X = 0) + P (X = 1)

= k + 2k = 3k = \(\frac{3}{6}=\frac{1}{2}\)

P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)

= k + 2k + 3k = 6k = \(\frac{6}{6}\) = 1 .

P (X ≥ 2) = 1 – P (X < 2) = 1 – \(\frac{1}{2}=\frac{1}{2}\)

Question 4.

A coin is tossed twice (or two coins are tossed simultaneously).

(i) Find the probability distribution of X, the number of heads. (NCERT)

(ii) If a random variable Y is defined on this sample space as the number of tails, then find the probability distribution of Y.

Answer:

(i) When a coin is tossed twice

Then sample space = {HH, HT, TH, TT} and all 4 events are equally likely.

Let random variable X be defined as number of heads

P (X = 0) = P (TT) = \(\frac{1}{4}\)

P(X= 1) = P(HT,TH) = \(\frac{2}{4}=\frac{1}{2}\)

P (X = 2) = P (HH) = \(\frac{1}{4}\)

Hence the required probability distribution of X is given below :

(ii) Let Y be the random variable which is the number of tails. So Y takes value 0, 1, 2

S = {HT, TH, TT, HH}

∴ P (Y = 0) = P (HH) = \(\frac{1}{4}\);

P (Y = 1) = P (HT, TH) = \(\frac{2}{4}\)

P (Y = 2) = P (TT) = \(\frac{1}{4}\)

∴ Probability distribution of Y is given by

Question 5.

A die is rolled. If a random variable X is defined as the number on the upper face, then find its probability distribution.

Answer:

When a die is rolled

Then sample space = {1, 2, 3, 4, 5, 6}

Let X be the random variable denotes the number on the upper face.

X takes values from 1 to 6.

P (X = 1) = \(\frac{1}{6}\) = P (X = 2) = P (X = 3)

= P (X = 4) = P (X = 5)

= P (X = 6)

Question 6.

An urn contains 5 white and 3 black balls. If two balls are drawn at random without replacement and X denotes the number of white balls drawn, then find its probability distribution.

Answer:

Number of white balls in a box = 5

No. of black balls in a box = 3

∴ Total no. of balls in a box = 5 + 3 = 8

Here, X be the random variable denote the number of white balls drawn in two draws of ball.

Then X can take values 0, 1, 2.

P (X = 0) = P (drawing no white balls)

= P (drawing 2 black balls)

= \(\frac{{ }^3 C_2}{{ }^8 C_2}=\frac{3 \times 2}{8 \times 7}=\frac{3}{28}\)

P (X = 1) = P (drawing one white ball and 1 black ball)

= \(\frac{{ }^5 \mathrm{C}_1 \times{ }^3 \mathrm{C}_1}{{ }^8 \mathrm{C}_2}=\frac{5 \times 3 \times 2}{8 \times 7}=\frac{15}{28}\)

P(X = 2) = P (drawing 2 white balls)

= \(\frac{{ }^5 \mathrm{C}_2}{{ }^8 \mathrm{C}_2}=\frac{5 \times 4}{8 \times 7}=\frac{5}{14}\)

required probability distribution of X is given by

Question 7.

Two cards are drawn simultaneously from a well-shuffled pack of 52 cards. Find the probability distribution of the number of jacks.

Answer:

Let X denotes the number of aces in 2 draws from a pack of 52 cards

∴ X can take values 0, 1,2.

P(X = 0) = probability of getting no ace = probability of getting two other cards

= \(\frac{{ }^{48} C_2}{{ }^{52} C_2}=\frac{48 \times 47}{52 \times 51}=\frac{12 \times 47}{13 \times 51}=\frac{188}{221}\)

P(X = 1) = probability of getting one ace and one other card

= \(\frac{{ }^4 C_1 \times{ }^{48} C_1}{{ }^{52} C_2}=\frac{4 \times 48 \times 2}{52 \times 51}=\frac{32}{221}\)

P(X = 2) = probability of getting two ace card

= \(\frac{{ }^4 C_2}{{ }^{52} C_2}=\frac{4 \times 3}{52 \times 51}=\frac{1}{221}\)

Thus the probability distribution of random variable X is given as under :

Question 8.

Four bad oranges are mixed accidentally with 16 good oranges. Find the probability distribution of the number of bad oranges in a draw of two oranges.

Answer:

Total no. of bad oranges = 4

Total no. of good oranges 16

∴ Total no. of oranges 4 ÷ 16 = 20

Let X denotes the number of bad oranges in a draw of 2 oranges. So X can take values 0, 1, 2.

2 oranges can be drawn simultaneously in 20C2 ways.

∴ P (X = 0) = P (drawing no bad orange) = P (drawing 2 good oranges)

= \(\frac{{ }^{16} C_2}{{ }^{20} C_2}=\frac{16 \times 15}{20 \times 19}=\frac{12}{19}\)

P (X = 1) = P (drawing one bad and one good orange)

= \(\frac{{ }^{16} C_2}{{ }^{20} C_2}=\frac{16 \times 15}{20 \times 19}=\frac{12}{19}\)

P (X = 2) = P (drawing 2 bad oranges)

= \(\frac{{ }^4 \mathrm{C}_2}{{ }^{20} \mathrm{C}_2}=\frac{4 \times 3}{20 \times 19}=\frac{3}{95}\)

The probability distribution of X is given below

Question 9.

Three balls are drawn one by one without replacement from a bag containing 5 white and 4 red balls. Find the probability distribution of the number of white balls drawn.

Answer:

Total no. of white balls = 5

Total no. of red balls = 4

∴ Total no. of balls = 5 + 4 = 9

Let X denote the random variable and denote the number of white balls drawn in a draw of three balls one by one without replacement.

∴ X can takes values 0, 1, 2, 3.

P(X = 0) = P (drawing no white ball)

= P (drawing three red balls)

= \(\frac{{ }^4 C_3}{{ }^9 C_3}=\frac{4 \times 3 \times 2}{9 \times 8 \times 7}=\frac{1}{21}\)

P(X = 1) = P (drawing 1 white ball and 2 black balls)

= \(\frac{{ }^5 \mathrm{C}_1 \times{ }^4 \mathrm{C}_2}{{ }^9 \mathrm{C}_3}=\frac{5 \times 4 \times 3 \times 6}{2 \times 9 \times 8 \times 7}=\frac{5}{14}\)

P(X = 2) = P (drawing 2 white balls and 1 black ball)

= \(\frac{{ }^5 \mathrm{C}_2 \times{ }^4 \mathrm{C}_1}{{ }^9 \mathrm{C}_3}=\frac{5 \times 4 \times 4 \times 6}{2 \times 9 \times 8 \times 7}=\frac{10}{21}\)

P(X = 3) = P (drawing 3 white balls) ‘

= \(\frac{{ }^5 C_3}{{ }^9 C_3}=\frac{5 \times 4 \times 6}{2 \times 9 \times 8 \times 7}=\frac{5}{42}\)

The probability distribution of random variable X is given by

Question 10.

A box contains 12 bulbs of which 3 are defective. A sample of 3 bulbs is selected from the ^^box. Let X denote the number of defective bulbs in the sample. Find the probability distribution of X.

Answer:

Total no. of bulbs = 12

no. of defective bulbs = 3

and no. of good bulbs = 12 – 3 = 9

Let X denote the no. of defective bulbs in a sample of 3 bulbs.

So X can take values 0, 1, 2, 3.

P (X = 0) = P (drawing no defective bulb = P (drawing 3 good bulbs)

= \(\frac{{ }^9 \mathrm{C}_3}{{ }^{12} \mathrm{C}_3}=\frac{9 \times 8 \times 7}{12 \times 11 \times 10}=\frac{42}{110}=\frac{21}{55}\)

P (X = 1) = P (drawing one defective and 2 non-defective bulbs)

= \(\frac{{ }^3 \mathrm{C}_1 \times{ }^9 \mathrm{C}_2}{{ }^{12} \mathrm{C}_3}=\frac{3 \times 9 \times 8 \times 6}{2 \times 12 \times 11 \times 10}=\frac{54}{110}=\frac{27}{55}\)

P (X = 2) = P (drawing two defective and one non-defective bulb)

= \(\frac{{ }^3 \mathrm{C}_2 \times{ }^9 \mathrm{C}_1}{{ }^{12} \mathrm{C}_3}=\frac{3 \times 9}{\frac{12 \times 11 \times 10}{6}}=\frac{3 \times 9 \times 6}{12 \times 11 \times 10}=\frac{27}{220}\)

P (X = 3) = P (drawing three defective bulbs)

= \(\frac{{ }^3 C_3}{{ }^{12} C_3}=\frac{1}{\frac{12 \times 11 \times 10}{6}}=\frac{1}{220}\)

Thus, the required probability distribution of X is given by

Question 11.

Four bad eggs are mixed with 10 good ones. If 3 eggs are drawn one by one without replacement, then find the probability distribution of the number of bad eggs drawn.

Answer:

Let X denotes the random variable denotes the no. of bad eggs.

∴ X takes values 0, 1, 2, 3.

No. of good eggs = 10

and No. of bad eggs = 4

P(X = 0) =P(No bad egg) = \(\frac{10}{14} \times \frac{9}{13} \times \frac{8}{12}=\frac{30}{91}\)

P (X = 1) = P (one bad egg = P (SFF, FSF, FFS)

= \(\frac{4}{14} \times \frac{10}{13} \times \frac{9}{12}+\frac{10}{14} \times \frac{4}{10} \times \frac{9}{12}+\frac{10}{14} \times \frac{9}{13} \times \frac{4}{12}\)

= \(\frac{3 \times 10 \times 9 \times 4}{14 \times 13 \times 12}=\frac{45}{91}\)

P (X = 2) = P (two badd eggs) = P (SSF, SFS, FSS)

= \(\frac{4}{14} \times \frac{3}{13} \times \frac{10}{12}+\frac{4}{14} \times \frac{10}{13} \times \frac{3}{12}+\frac{10}{14} \times \frac{4}{13} \times \frac{3}{12}\)

= 3 × \(\frac{4 \times 3 \times 10}{14 \times 13 \times 12}=\frac{15}{91}\)

P (X = 3) = P (three bad eggs) = \(\frac{4}{14} \times \frac{3}{13} \times \frac{2}{12}=\frac{1}{91}\)

Required probability distribution of X is given by

Question 12.

Four cards are drawn simultaneously from a well-shuffled pack of 52 cards. If X denotes ‘ the number of aces drawn, then find the probability distribution of X.

Answer:

Let X denotes the number of aces in a sample of 4 cards drawn from a well shuffled pack of 52 cards.

So X can take values 0, 1,2, 3, 4.

P(X = 0) = probability of getting no ace

= \(\frac{{ }^{48} C_4}{{ }^{52} C_4}\)

P(X = 1) = probability of getting one ace and 3 other cards

= \(\frac{{ }^4 C_1 \times{ }^{48} C_3}{{ }^{52} C_4}\)

P(X = 2) – prob. of getting two aces and 2 other cards

= \(\frac{{ }^4 C_2 \times{ }^{48} C_2}{{ }^{52} C_4}\)

P(X = 3) = prob. of getting three aces and one other card

= \(\frac{{ }^4 C_3 \times{ }^{48} C_1}{{ }^{52} C_4}\)

P(X = 4) = probability of getting four aces

= \(\frac{{ }^4 C_4}{{ }^{52} C_4}\)

The probability distribution of random variable X is given as under: