Students often turn to ML Aggarwal Class 12 Solutions ISC Chapter 10 Probability Ex 10.10 to clarify doubts and improve problem-solving skills.

## ML Aggarwal Class 12 Maths Solutions Section A Chapter 10 Probability Ex 10.10

Question 1.

Three letters are written to three persons and an envelope is addressed to each of them, the letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the probability that atleast one letter is in its proper envelope.

Answer:

Aliter:

Let the three envelops be donated by E_{1}, E_{2} and E_{3} and three letters be denoted by L_{1}, L_{2}, and L_{3}. For getting atleast one letter should be in right envelope following possibilities arises.

(i) 1 letter in correct envelope and other 2 in wrong envelope

(a) E_{1}L_{1}, E_{2}L_{2}, E_{3}L_{3}

(b) E_{1}L_{3}, E_{2}L_{2}, E_{3}L_{1}

(c) E_{1} L_{2}, E_{2}L_{1}, E_{3}L_{3}

(ii) two letters in correct envelope i.e. (E_{1}L_{1}, E_{2}L_{2}, E_{3}L_{3}}

Thus total no. of favourable cases = 3 + 1 = 4 and

Total no. of possibilities = 3! = 6

∴ required prob. = \(\frac{4}{6}=\frac{2}{3}\)

Question 2.

A die is so loaded that the probability of face ¡ is proportional to ¡ (where 1 1, 2, …, 6). What Is the probability of an even number occurring when the die is rolled?

Answer:

Since probability of fare jis proportional to I.

P (face i) = k i; where k = constant of proportionality

P(face 1) = k; P(2) = 2k; P(3) = 3k

P (4) = 4k ; P (5) = 5k and P (6) = 6k

Since ΣP_{i} = 1

⇒ k + 2k + 3k + 4k + 5k + 6k = 1

⇒ 21k = 1

⇒ k = \(\frac{1}{21}\)

∴ Prob. (even number) = P (2) + P (4) + P (6)

= 2k + 4k + 6k = 12k

= \(\frac{12}{21}=\frac{4}{7}\)

Question 3.

A coin is tossed n times. What is the chance that the head will present itself an odd number of times ?

Answer:

Since a coin is tossed n times

∴ Total no. of outcomes = 2^{n}

∴ Total no. of favourable outcomes = ^{n}C_{1} + ^{n}C_{3} + ….

= Sum of odd binomial coefficients

= 2^{n-1}

required probability = \(\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}=\frac{2^{n-1}}{2^n}=\frac{1}{2}\)

Question 4.

If A, B and C are the events associated with a random experiment such that P (exactly one of the events A or B) = P (exactly one of B or C) = P (exactly one of C or A) = p and P (all of three events occurring simultaneously) = p , then find the probability of occurring of atleast of one of the three events A, B and C.

Answer:

Given P (exactly one of the events A or B) = p (Ā ∩ B) + P (A ∩ B̄) = p (given)

⇒ P (A) – P (A ∩ B) + P (B) – P (A ∩ B) = p

⇒ P (A) + P (B) – 2 P (A n B) = p …(1)

P (exactly one of the events B or C) = p

⇒ P(B̄ ∩ C) + P(B ∩ C̄) = p

⇒ P (B) + P (C)- 2P (B ∩ C) = p …(2)

Similarly P (C) + P (A) – 2P (A ∩ C) =p …(3)

also given P (A ∩ B ∩ C) = p2 …(4)

We want to find probability of occurring of atleast of one of three events A, B and C i.e. P(A ∪ B ∪ C)

SinceP(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(C ∩ A) + P(A ∩ B ∩ C)

= \(\frac{1}{2}\) [2P (A) + 2P (B) + 2P (C) – 2P (A ∩ B) – 2 P (B ∩ C) – 2P (C ∩ A) + 2P (A ∩ B ∩ C)]

= \(\frac{1}{2}\)[{P(A) + P(B) – 2P(A ∩ B)} + {P(B) + P(C) – 2P(B ∩ C)} + P(C) + P(A) – 2P(A ∩ C) + 2P(A ∩ B ∩ C)}]

= \(\frac{1}{2}\)[p + p + p + 2p^{2}] [Using (1), (2), (3) and (4)]

= \(\frac{1}{2}\)(3p + 2p^{2})

Question 5.

E and F are two independent events. The probability that E and F both occur is \(\frac{1}{12}\) and the probability neither E nor F occur is \(\frac{1}{2}\). Find P (E) and P (F).

Answer:

Given P(E ∩ F) = \(\frac{1}{12}\) …………..(1)

and P(Ē ∩ F̄) = \(\frac{1}{2}\) ……………(2)

From (2); P\((\overline{E \cup F})\) = \(\frac{1}{2}\)

⇒ 1 – P(E ∪ F) = \(\frac{1}{2}\)

⇒ P(E ∪ F) = \(\frac{1}{2}\)

⇒ P(E) + P(F) = \(\frac{1}{2}\)

⇒ P(E) + P(F) – P(E ∩ F) = \(\frac{1}{2}\)

⇒ P(E) + P(F) = \(\frac{1}{2}+\frac{1}{12}=\frac{6+1}{12}=\frac{7}{12}\) ……(3) [Using (1)]

Since E and F are independent events

∴ from (1); P(E) P(F) = \(\frac{1}{12}\) ……(4)

from (3); we have P(F)[\(\frac{7}{12}\) – P(E)] = \(\frac{1}{12}\)

Question 6.

The probabilities that a student passes in Mathematics, Physics and Chemistry are m, p and c respectively. Of these subjects, the student has 75% chance of passing in atleast one subject, a 50% chance of passing in atleast two subjects and a 40% chance of passing in exactly two subjects. Find the value of

(i) m +p + c

(ii) mpc

Answer:

Let A, B and C be the events that the student passed in Mathematics, Physics and Chemistry.

P (A) = m ;P (B) = p ; P (C) = c

given P(A ∪ B ∪ C) = 75% = \(\frac{3}{4}\)

Also, P(A ∩ B) + P(B ∩ C) + P(C ∩ A) – 2P(A ∩ B ∩ C) = \(\frac{1}{2}\) ..(1)

P(A ∩ B) + P(B ∩ C) + P(C ∩ A) – 3P(A ∩ B ∩ C) = \(\frac{40}{100}=\frac{2}{5}\) ……(2)

We know that

P (A ∪ B ∪ C) = P (A) + P (B) + P (C) – P (A ∩ C) – P (B ∩ C) – P (C ∩ A) + P (A ∩ B ∩ C)

\(\frac{3}{4}\) = P (A) + P (B) + P (C) – [\(\frac{1}{2}\) + 2P (A ∩ B ∩ C)] + P(A ∩ B ∩ C)

\(\frac{3}{4}\) = P (A) + P (B) + P (C) – \(\frac{1}{2}\) – P (A ∩ B ∩ C)] ……(3)

eqn. (1) – eqn. (2)gives P(A ∩ B ∩ C) = \(\frac{1}{2}-\frac{2}{5}=\frac{1}{10}\)

from(3);P(A) + P(B) + P(C) = \(\frac{3}{4}+\frac{1}{2}+\frac{1}{10}=\frac{15+10+2}{20}=\frac{27}{20}\)

⇒ m + p + c = \(\frac{27}{20}\)

Since A, B, C are independent events and P(A ∩ B ∩ C) = \(\frac{1}{10}\)

⇒ P (A) . P (B) P (C) = \(\frac{1}{10}\)

⇒ mpc = \(\frac{1}{10}\)

Question 7.

A box contains 6 red, 4 white and 5 black balls. A person draws 4 balls from the box at random. Find the probability that among the balls drawn there is atleast one ball of each colour.

Answer:

Given Total no. of red balls = 6

Total no. of white balls = 4

Total no. of black balls = 5

∴ Total no. of balls = 6 + 4 + 5= 15

Thus Total no. of ways of drawing 4 balls out of 15 be 15C4

∴ required probability = P (drawing 2 red balls one white and one black ball) . + P (drawing 1 red ball, 2 white and 1 black ball) + P (drawing 1 red, 1 white, 2 black balls)

Question 8.

Munna enjoys flying his kite. On any given day, the probability that there is a good wind is \(\frac{3}{4}\). If there is a good wind, the probability that the kite will fly is \(\frac{5}{8}\). If there is not a good wind, the probability that the kite will fly is \(\frac{1}{16}\). further, if the kite flies, the probability that it sticks ¡n a tree is \(\frac{1}{2}\). Draw a probability tree diagram for this and hence calculate the following probabilities :

(i) that there is a good wind and the kite flies.

(ii) that whatever the wind, the kite does not fly.

(iii) that whatever the wind, the kite sticks in a tree.

Answer:

Let us define the events K and W as follows :

K : kite will fly

W : there is a good wind

Given P (W) = P (there is a good wind)

P (W̄) = P (there is not a good wind)

= 1 – \(\frac{3}{4}=\frac{1}{4}\)

P(K/W) = \(\frac{5}{8}\),P(K̄/W) = 1 – \(\frac{5}{8}=\frac{3}{8}\)

P(K/W̄) = \(\frac{1}{16}\), P(K̄/W̄) = 1 – \(\frac{1}{16}=\frac{15}{16}\)

Let S be the event that the kite sticks in tree

∴ P(S/K) = \(\frac{1}{2}\)

(i) required probability = P (W ∩ K)

= P(K/W)P(W) = \(\frac{5}{8} \times \frac{3}{4}=\frac{15}{32}\)

(ii) required probability = P(K̄)

P(W)P(K̄/W) + P(W̄)P(K̄/W)

= \(\frac{3}{4} \times \frac{3}{8}+\frac{1}{4} \times \frac{15}{16}=\frac{18+15}{64}=\frac{33}{64}\)

(iii) required probability = P (W) P (K/W)

P(S/K)+ P (W̄) P (K/W̄) P (S/K)

= \(\frac{3}{4} \times \frac{5}{8} \times \frac{1}{2}+\frac{1}{4} \times \frac{1}{16} \times \frac{1}{2}\)

= \(\frac{30+1}{128}=\frac{31}{128}\)

Question 9.

Two bags P and Q contain 4 white, 3 black balls and 2 white, 2 black balls respectively. From bag P two balls are transferred to bag Q. Find the probability of drawing 1 white and 1 black ball from bag Q.

Answer:

Given bag P contains 4 white and 3 black balls

bag Q contains 2 white and 2 black balls

Let us define the events E_{1}, E_{2}, E_{3} and A as follows:

E_{1} : transferring 2 white balls from bag P to bag Q

E_{2} : transferring 2 black balls from bag P to bag Q

E_{3} : transferring 1 white and 1 black from P to Q

A : drawing 1 white and 1 black ball from bag Q.

P (A/E_{1}) = P (drawing 1 white and 1 black ball from bag Q after transferring 2 white balls from P to Q)

= \(\frac{{ }^2 \mathrm{C}_1 \times{ }^4 \mathrm{C}_1}{{ }^6 \mathrm{C}_2}=\frac{4 \times 2 \times 2}{6 \times 5}=\frac{16}{30}=\frac{8}{15}\)

P (A/E_{2}) = P (drawing 1 white and 1 black ball from Q after transferring 2 black balls from P to Q)

= \(\frac{{ }^2 \mathrm{C}_1 \times{ }^4 \mathrm{C}_1}{{ }^6 \mathrm{C}_2}=\frac{2 \times 4 \times 2}{6 \times 5}=\frac{16}{30}=\frac{8}{15}\)

P (A/E_{3}) = P (drawing 1 white and 1 black ball from Q after transferring 1 white and 1 black ball from P to Q)

= \(\frac{{ }^3 \mathrm{C}_1 \times{ }^3 \mathrm{C}_1}{{ }^6 \mathrm{C}_2}=\frac{3 \times 3 \times 2}{6 \times 5}=\frac{18}{30}=\frac{3}{5}\)

Since E_{1}, E_{2}, E_{3} are mutually exclusive and exhaustive events.

Then required probability = P (A)

= P (E_{1}) P (A/E_{1}) + P (E_{2}) P (A/E_{2}) + P (E_{3}) P (A/E_{3})

= \(\frac{16+8+36}{105}=\frac{60}{105}=\frac{4}{7}\)

= \(\frac{16+8+36}{105}=\frac{60}{105}=\frac{4}{7}\)

Question 10.

A football coach knows that his team will play 40% of its games on artificial turf this season. He also knows that a player’s chances of incurring a knee injury are 50% higher if he is playing on artificial turf instead of grass. If a player’s probability of knee injury on artificial turf is 0-42, what is the probability that

(i) a randomly selected player incurs a knee injury ?

(ii) a randomly selected player with a knee injury incurred the injury playing on grass ?

Answer:

Let us define the events E_{1}, E_{2} and A are as follows:

E_{1} : playing on artificial turf

E_{2} : playing on grass

A : selected player incurs a knee injury

Then p(E_{1}) = \(\frac{40}{100}\); p(E_{2}) = \(\frac{60}{100}\)

E_{1} and E_{2} are mutually exclusive and exhaustive events.

Let the player has knee injury while playing on grass has probability x

Then according to given,

x + 50% of x = 0.42

⇒ \(\frac{3 x}{2}\) = 0.42

⇒ x = \(\frac{0.42 \times 2}{3}\) = 0.28

p (A/E_{1}) = 0.42 and p(A/E_{2}) = 0.28

(i) required probability = p (A)

= p(E_{1})p(A/E_{2}) + p(E_{2})p(A/E_{2})

[using law of Total probability]

= \(\frac{40}{100}\) × 0.42 + \(\frac{60}{100}\) × 0.28

= 0.1680 + 0.1680

= 0.3360

(ii) We want to find p (E2/A)

Then by Baye’s Theorem, we have

Question 11.

An urn contains 5 red and 4 white balls. A ball is drawn at random, its colour is noted and is returned to the urn along with two additional balls of the same colour and then a ball is drawn at random.

(i) What is the probability that the second ball drawn is red ?

(ii) If a red ball has been drawn in the second draw, then find the probability that a red ball was drawn in the first draw.

Answer:

Let us define the events E_{1}, E_{2} and A as follows:

E_{1} : A red ball is drawn in 1st draw

E_{2} : A white ball is drawn in 1st draw

A : The ball drawn be of red colour in 2nd draw

Then p(E_{1}) = \(\frac{5}{9}\);

p(E_{2}) = \(\frac{4}{9}\)

So E_{1} and E_{2} are mutually exclusive and exhaustive events.

p (A/E_{1}) = p (drawing a red ball in 2nd draw when a red ball is already drawn in 1st draw)

= \(\frac{7}{7+4}=\frac{7}{11}\)

p (A/E_{2}) = prob. of drawing a red ball in 2nd draw when a white ball is already drawn in first draw

= \(\frac{7}{7+4}=\frac{7}{11}\)

Then by law of Total probability : we have

P(A) = P(E_{1})P(A/E_{1}) + P(E_{2}) P(A/E_{2})

= \(\frac{5}{9} \times \frac{7}{11}+\frac{4}{9} \times \frac{5}{11}=\frac{35+20}{99}=\frac{55}{99}=\frac{5}{9}\)

(ii) We want to find P(E_{1}/A)

By Baye’s Theorem, we have

Question 12.

There are two biased coins. Coin 1 turns up heads only 40% of the time while coin 2 turns up heads 70% of the times.

(i) A coin is picked up. What is the probability that it is of type 1 ?

(ii) A coin is picked up. It is tossed once and conies up heads. What is the probability that it is of type 1 ?

(iii) A coin is picked up. It is tossed twice and conies up heads both times. What is the probability that it is of type 1 ?

Answer:

Let us define the events E_{1}, E_{2} and A are as follows:

E_{1}: coin 1 is selected

E_{2} : coin 2 is selected A : coin comes up head.

Here P (E_{1}) = \(\frac{1}{2}\) = P(E_{2})

and P (A/E_{1}) = \(\frac{2}{5}\) and P (A/E_{2}) = \(\frac{7}{10}\)

(i) required probability = \(\frac{1}{2}\)

(ii) We want to find P (E_{1}/A)

(iii) When a coin tossed twice

Question 13.

Bag I contains 4 red and 3 white balls ; bag II contains 2 red and 2 white bafls. Two balls are transferred from bag I to bag II and then two balls are drawn from bag II. If the balls drawn from bag II are one red and one white, then find the probability that 2 red balls were transferred from bag I to bag II.

Answer:

Let us define the events E_{1}, E_{2}, E_{3} and A are as follows:

E_{1} : Two red balls are transferred from bag I to bag II

E_{2} : Two white balls are transferred from bag I to bag II

E_{3} : one red and one white ball is transferred from bag I to bag II

A : one red ball and one white ball drawn from bag II

∴ E_{1}, E_{2} and E_{3} are mutually exclusive and exhaustive events.

P (A/E_{1}) = P (drawing one red ball and one white ball from bag II when two red balls are already transferred from I to II)

= \(\frac{{ }^4 \mathrm{C}_1 \times{ }^2 \mathrm{C}_1}{{ }^6 \mathrm{C}_2}=\frac{4 \times 2 \times 2}{6 \times 5}=\frac{16}{30}=\frac{8}{15}\)

P (A/E_{2}) = P (drawing one red and one white ball from bag II when two white balls are already transferred from I to II)

= \(\frac{{ }^2 \mathrm{C}_1 \times{ }^4 \mathrm{C}_1}{{ }^6 \mathrm{C}_2}=\frac{2 \times 4 \times 2}{6 \times 5}=\frac{8}{15}\)

P (A/E_{3}) = P (drawing one red and one white ball from bag II when one red and one white ball are already transferred from bag I to bag II)

= \(\frac{{ }^3 \mathrm{C}_1 \times{ }^3 \mathrm{C}_1}{{ }^6 \mathrm{C}_2}=\frac{3 \times 3 \times 2}{6 \times 5}=\frac{18}{30}=\frac{3}{5}\)

We want to find P (E(/A)

Then by Baye’s Theorem, we have

Question 14.

In a multiple choice question, there are four alternative answers, of which one or more are correct. A candidate will get marks in the question only if he ticks all the correct answers. The candidate decides to tick answers at random. If he is allowed upto three chances to answer the question, then find the probability that he will get marks in the question.

Answer:

Here Total no. of possible tickings = 24 – 1 = 15 as the candidate ticks atleast one choice.

Out of 15, only one combination is correct.

∴ p = prob. of success = \(\frac{1}{15}\)

q = 1 – p = 1 – \(\frac{1}{15}=\frac{14}{15}\)

and here n = 3

Required probability that he will get marks in the question

P(X ≥ 1) = 1 – P(X = 0) …………(1)

By binomial distribution, we have

Question 15.

What is the least number of times a pair of dice must be rolled so that the probability of getting a sum 12 atleast once shall be more than \(\frac{1}{2}\)?

Answer:

Let us consider the required no. of times, a pair of dice must be rolled be n so that Probability of getting a sum 12 atleast once shall be more than \(\frac{1}{2}\).

Le. P (X ≥ 1) > \(\frac{1}{2}\)

⇒ 1 – P (X = 0) > \(\frac{1}{2}\) …(1)

When a pair of dice are thrown Then total no. of outcomes = 62 For getting a sum of 12, favourable outcomes = {6, 6}

∴ p = prob. of getting sum 12 = \(\frac{1}{36}\)

∴ q = 1 – p = 1 – \(\frac{1}{36}=\frac{35}{36}\)

Then by binomial distribution, we have

eqn. (2) is satisfied when n is atleast 25. Thus, the pair of dice must be rolled atleast 25 times.

Question 16.

Find the mean (or expectation) of a discrete random variable X whose probability function is given by f(x) = \(\left(\frac{1}{2}\right)^x\) where x = 1, 2, 3,…

Answer:

The given function is f(x) = \(\left(\frac{1}{2}\right)^x\) where x = 1, 2, 3, ….

Required mean (or Expectation)

= E(X) = Σxf(x)

= 1 × \(\frac{1}{2}\) + 2 × \(\frac{1}{4}\) + 3 × \(\frac{1}{8}\) + 4 × \(\frac{1}{16}\) ……….(1)

Let S = \(\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}\) ….(2)

Multiply eqn. (2) by \(\frac{1}{2}\) we have

\(\frac{1}{2}\) S = \(\frac{1}{4}+\frac{2}{8}+\frac{3}{16}\) + ……… ….(3)

eqn. (2) – eqn. (3); we have S 1 1 1 1

\(\frac{S}{2}=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\) + ………….∞

= \(\frac{\frac{1}{2}}{1-\frac{1}{2}}=\frac{\frac{1}{2}}{\frac{1}{2}}\) = 1

S = 2

∴ from (1); required mean = 2

Question 17.

A random variable X can take all non-negative integral values and the probability that X takes the value r is proportional to ar (0 < a < 1). Find P(X = 0).

Answer:

Given P (X = r) ∝ αr (0 < α < 1)

⇒ P(r) = kαr, where k be the constant of proportionality

where r = 0, 1, 2, 3, …

Since Σp_{i} = 1

⇒ k (α^{0} + α^{1} + α^{2} …. ∞) = 1

⇒ k\(\left(\frac{1}{1-\alpha}\right)\) = 1

⇒ k = (1 – α)

.-. P (X = r) = (1 – α) α^{r}

Thus, P (X = 0) = (1 – α) α^{0} = 1 – α