NCERT Exemplar: Probability | Mathematics (Maths) Class 12 - JEE PDF Download

SHORT ANSWER TYPE QUESTIONS

Q.1. For a loaded die, the probabilities of outcomes are given as under:
P(1) = P(2) = 0.2, P(3) = P(5) = P(6) = 0.1 and P(4) = 0.3.
The die is thrown two times. Let A and B be the events, ‘same number each time’, and ‘a total score is 10 or more’, respectively. Determine whether or not A and B are independent.
Ans. A loaded die is thrown such that
P(1) = P(2) = 0.2,
P(3) = P(5) = P(6) = 0.1
and P(4) = 0.3
die is thrown two times.
Also given that:
A = Same number each time and
B = Total score is 10 or more.
So, P(A) - [P(1,1) + P(2, 2) + P(3, 3) + P(4, 4) + P(5, 5) + P(6, 6)]
= P(1).P(1) + P(2).P(2) + P(3).P(3) + P(4).P(4) + P(5).P(5) + P(6).P(6)
= 0.2 x 0.2 + 0.2 x 0.2 + 0.1 x 0.1 + 0.3 x 0.3 + 0.1 x 0.1 + 0.1 x 0.1
= 0.04 + 0.04 + 0.01 + 0.09 + 0.01 + 0.01 = 0.20
Now B = ((4, 6), (6, 4), (5, 5), (5, 6), (6, 5), (6, 6)]
P(B) = |P(4).P(6) + P(6).P(4) > P(5).P(5) + P(5).P(6) + P(6).P(5) + P(6).P(6)
= 0.3 x 0.1 +0.1 x 0.3+ 0.1 x 0.1 +0.1 x 0.1 +0.1 x 0.1 + 0.1 x 0.1
= 0.03 + 0.03 + 0.01 + 0.01 + 0.01 + 0.01 = 0.10
A and B both events will be independent if
P(A ∩ B) = P(A).P(B)   ...(i)
Here, (A ∩ B) = {(5, 5), (6,6)}
∴ P(A ∩ B) = P(5, 5) + P(6, 6) = P(5).P(5) + P(6).P(6)
= 0.1 x 0.1 + 0.1 x 0.1 = 0.02
From eq. (i) we get
0.0 2 = 0. 20 x 0 .1 0
0.02 = 0.02
Hence, A and B are independent events.

Q.2. Refer to Exercise 1 above. If the die were fair, determine whether or not the events A and B are independent.
Ans. We have A = { (1.1). (2. 2), (3. 3). (4.4). (5, 5), (6.6)}
∴ n(A) = 6 and n{S) = 6² = 36
∴
and B = {(4.6).(6,4),(5.5).(6.5).(5,6),(6.6)}
⇒ n(B) = 6
∴
Also, (A ∩ B) = {(5, 5), (6,6)}
∴
Also,
Thus, P(A ∩ B) ≠ P(A).P(B)
So. A and B are not independent events.

Q.3. The probability that at least one of the two events A and B occurs is 0.6. If A and B occur simultaneously with probability 0.3, evaluate.
Ans. According to the question,
P(A ∪ B ) = 0.6 and P{A ∩ B) = 0.3
Now P(A ∪ B) = P(A) + P(B) - P(A ∩ B)


Q.4. A bag contains 5 red marbles and 3 black marbles. Three marbles are drawn one by one without replacement. What is the probability that at least one of the three marbles drawn be black, if the first marble is red?
Ans. Bag contains 5 red marbles and 3 black marbles.
For atleast one of the three marbles drawn be black, if the first marble is red, then the following situations are possible
Event E₁ = Second marble is black and third is red
Event E₂ = Second marble is black and third is also black (E₂)
Event E₃ = Second marble is red and third is black (E₃)
Let Event R_i = drawing red marble in i^th draw
Event B_i = drawing black marble in i^th draw



Q.5. Two dice are thrown together and the total score is noted. The events E, F and G are ‘a total of 4’, ‘a total of 9 or more’, and ‘a total divisible by 5’, respectively. Calculate P(E), P(F) and P(G) and decide which pairs of events, if any, are independent.
Ans. Two dice are thrown together i.e.,
∴ n(S) = 36, where S is the sample space.
Event ‘E' is ‘a total o f 4 ’
E = {( 2, 2), ( 3, 1), ( 1, 3)}x
Event ‘F' is la total of 9 or more’
∴ F= {(3,6), (6,3), (4, 5), (5,4). (4, 6). (6,4), (5, 5), (5,6), (6, 5), (6,6)}
Event 'G' is ‘a total divisible by 5’
∴ G - {(1,4), (4, 1), (2,3), (3,2), (4,6), (6,4), (5, 5)}
Here, (E ∩ F) = φ and (E ∩ G) = φ
Also, ( F ∩ G ) = {(4, 6), ( 6 .4 ). (5, 5)}


and

So,
Hence, there is no pair which is independent.

Q.6. Explain why the experiment of tossing a coin three times is said to have binomial distribution.
Ans. We know that, in a Binomial distribution,
(i) There are 2 outcomes for each trial
(ii) There is a fixed number of trials
(iii) The probability of success must be the same for all the trials.
When coin is tossed, possible outcomes are Head and Tail. Since coin is tossed three times, we have fixed number of trials. Also probability of Head and Tail in each trial is 1/2. Thus given experiment is said to have binomial distribution.

Q.7. A and B are two events such that P(A) = 1/2, P(B) = 1/3 and P(A ∩ B) = 1/4.
Find:
(1) P(A|B)
(2) P(B|A)
(3) P(A'|B)
(4) P(A'|B')
Ans.


Q.8. Three events A, B and C have probabilities 2/5, 1/3 and 1/2, respectively. Given that P(A ∩ C) = 1/5 and P(B ∩ C) = 1/4, find the values of P(C | B) and P(A'∩ C').
Ans.


Q.9. Let E₁ and E₂ be two independent events such that p(E₁) = p₁ and P(E₂) = p₂. Describe in words of the events whose probabilities are:
(1) p₁ p₂
(2) (1–p₁) p₂
(3) 1–(1–p₁)(1–p₂)
(4) p₁ + p₂ – 2p₁p₂
Ans. P(E₁) - P1 and P(E₂) = p₂
(i) 
So, E₁ and E₂ occur simultaneously.
(ii) 
So, E₁ does not occur but E₂ occurs.
(iii)

So, either E₁, or E₂ or both E₁ , and E₂ occurs.
(iv)

So, either E₁ or E₂ occurs but not both.

Q.10. A discrete random variable X has the probability distribution given as below:

(1) Find the value of k
(2) Determine the mean of the distribution.
Ans. For a probability distribution, we know that if P_i ≥ 0
(i)
⇒ k + k² + 2k² + k = 1
⇒ 3k² + 2k - 1 = 0
⇒ 3k² + 3k - k - 1 = 0
⇒ 3k(k + 1) - 1(k + 1) = 0
⇒ (3k - 1)(k + 1) = 0
∴ k = 1/3 and k = – 1
But k ≥ 0
∴ k = 1/3
(ii) Mean of the distribution



Q.11. Prove that
(1) P(A) = P(A ∩ B) +
(2) P(A ∪ B) = P(A ∩ B) +
Ans. 
(i) To prove: P(A) = P(A ∩ B) +
R.H.S. = P(A ∩ B) +
= P(A).P(B) + P(A) .= P(A) [P(B) +]
= P(A) . 1
= P(A) = L.H.S. Hence proved.
(ii) To prove: P(A ∪ B) = P(A ∩ B) + P(A + ∩ B)
R.H.S. = P(A) . P(B) + P(A) .+ . P(B)
= P(A).P(B) + P(A)[1 – P(B)] + [1 – P(A)].P(B)
= P(A).P(B) + P(A) – P(A).P(B) + P(B) – P(A).P(B)
= P(A) + P(B) – P(A ∩ B)
= P(A ∪ B) = L.H.S. Hence proved.

Q.12. If X is the number of tails in three tosses of a coin, determine the standard deviation of X.
Ans.Given that: X = 0, 1, 2, 3

Probability distribution table is:


we know that  Var(X) =
∴ Standard deviation =

Q.13. In a dice game, a player pays a stake of Re 1 for each throw of a die. She receives Rs 5 if the die shows a 3, Rs 2 if the die shows a 1 or 6, and nothing otherwise. What is the player’s expected profit per throw over a long series of throws?
Ans. Let X be the random variable of profit per throw.

Since, she loses Rs.1 for getting any of 2, 4, 5.
So,

Player ’s expected profit =


Q.14. Three dice are thrown at the sametime. Find the probability of getting three two’s, if it is known that the sum of the numbers on the dice was six.
Ans.The dice is thrown three times
∴ Sample space n(S) = (6)³ = 216
Let E₁ be the event when the sum of numbers on the dice was six and E₂ be the event when three two’s occur.
⇒ E₁ = {(1, 1, 4), (1, 2, 3), (1, 3, 2), (1, 4, 1), (2, 1, 3), (2, 2, 2), (2, 3, 1), (3, 1, 2), (3, 2, 1), (4,1, 1)} [∴ E₂ = {2, 2, 2}]
⇒ n(E₁) = 10 and n(E₂) = 1


Q.15. Suppose 10,000 tickets are sold in a lottery each for Re 1. First prize is of Rs 3000 and the second prize is of Rs. 2000. There are three third prizes of Rs. 500 each. If you buy one ticket, what is your expectation.
Ans. Let X be the random variable where X = 0, 500, 2000 and 3000


Hence, expectation is Rs.0.65.

Q.16. A bag contains 4 white and 5 black balls. Another bag contains 9 white and 7 black balls. A ball is transferred from the first bag to the second and then a ball is drawn at random from the second bag. Find the probability that the ball drawn is white.
Ans. Let W₁ and W₂ be two bags containing (4 W, 5 B) and (9 W, 7 B) balls respectively. Let E₁ be the event that the transferred ball from the bag W₁ to W₂ is white and E₂ the event that the transferred ball is black.
And E be the event that the ball drawn from the second bag is white.


Hence, the required probability is 5/9.

Q.17. Bag I contains 3 black and 2 white balls, Bag II contains 2 black and 4 white balls. A bag and a ball is selected at random. Determine the probability of selecting a black ball.
Ans. Given that bag I = {3 B, 2 W}
and bag II = {2 B, 4 W}
Let E₁ = The event that bag I is selected
E₂ = The event that bag II is selected
and E = The event that a black ball is selected

Hence, the required probability is 7/15.

Q.18. A box has 5 blue and 4 red balls. One ball is drawn at random and not replaced. Its colour is also not noted. Then another ball is drawn at random. What is the probability of second ball being blue?
Ans. Given that the box has 5 blue and 4 red balls.
Let E₁ be the event that first ball drawn is blue
E₂ be the event that first ball drawn is red
and E is the event that second ball drawn is blue.
∴ P(E) = P(E₁) . P (E/E₁) + P(E₂) . P(E/E₂)

Hence, the required probability is 5/9.

Q.19. Four cards are successively drawn without replacement from a deck of 52 playing cards. What is the probability that all the four cards are kings?
Ans. Let E₁, E₂, E₃ and E₄ be the events that first, second, third and fourth card is King respectively.
∴ P(E₁ ∩ E₂ ∩ E₃ ∩ E₄)

Hence, the required probability is 1/27075.

Q.20. A die is thrown 5 times. Find the probability that an odd number will come up exactly three times.
Ans. Here,
and n = 5

Hence, the required probability is 5/16.

Q.21. Ten coins are tossed. What is the probability of getting at least 8 heads?
Ans.


Hence, the required probability is 7/128.

Q.22. The probability of a man hitting a target is 0.25. He shoots 7 times. What is the probability of his hitting at least twice?
Ans.Here n  = 7, p = 0.25 =

Hence, the required probability is 4547/8192.

Q.23. A lot of 100 watches is known to have 10 defective watches. If 8 watches are selected (one by one with replacement) at random, what is the probability that there will be at least one defective watch?
Ans. Probability of defective watch out of 100 watches = 10/100 = 1/10.
Here, n  = 8, p = 1/10 and q = 1 - 1/10 = 9/10 and r ≥ 1
P(X ≥ 1) = 1 - P(x = 0) =
Hence, the required probability is .

Q.24. Consider the probability distribution of a random variable X:

Calculate (1)
(2) Variance of X.
Ans. Here, we have

We know that: Var(X) = E(X²) – [E(X)]²
where E(X) = and E(X²) =
∴ E(X) = 0 × 0.1 + 1 × 0.25 + 2 × 0.3 + 3 × 0.2 + 4 × 0.15
= 0 + 0.25 + 0.6 + 0.6 + 0.6 = 2.05
E(X²) = 0 × 0.1 + 1 × 0.25 + 4 × 0.3 + 9 × 0.2 + 16 × 0.15
= 0 + 0.25 + 1.2 + 1.8 + 2.40 = 5.65

(ii) Var(X) = 1.4475

Q.25. The probability distribution of a random variable X is given below:

(1) Determine the value of k.
(2) Determine P(X ≤ 2) and P(X > 2)
(3) Find P(X ≤ 2) + P (X > 2).
Ans.(1) We know that P(0) + P(1) + P(2) + P(3) = 1

∴ k = 8/15
(2)

(3)


Q.26. For the following probability distribution determine standard deviation of the random variable X.

Ans.We know that: Standard deviation (S.D.) =


Q.27. A biased die is such that P(4) = 1/10 and other scores being equally likely. The die is tossed twice. If X is the ‘number of fours seen’, find the variance of the random variable X.
Ans. Here, random variable X = 0, 1, 2



We know that V(X) = E(X²) – [E(X)]²

∴
Hence, the required variance = 0.18.

Q.28. A die is thrown three times. Let X be ‘the number of twos seen’. Find the expectation of X.
Ans. Here, we have X = 0, 1, 2, 3 [∴ die is thrown 3 times]
and p = 1/6, q = 5/6
∴P(X = 0) = P(not 2) . P(not 2) . P(not 2) =
P(X = 1) = P(2) . P(not 2) . P(not 2) + P(not 2) . P(2) . P(not 2) + P(not 2) . P(not 2) . P(2)

Now

Hence, the required expectation is 1/2.

Q.29. Two biased dice are thrown together. For the first die P(6) = 1/2, the other scores being equally likely while for the second die, P(1) = 2/5 and the other scores are equally likely. Find the probability distribution of ‘the number of ones seen’.
Ans. Given that: for the first die, P(6) = 1/2 and= 1- 1/2 = 1/2

For the second die, P(1) = 2/5 and = 1- 2 =3/5
Let X be the number of one’s seen
∴ X = 0, 1, 2

Hence, the required probability distribution is


Q.30. Two probability distributions of the discrete random variable X and Y are given below.

Prove that E(Y²) = 2 E(X).
Ans. First probability distribution is given by

We know that, E(X) =

For the second probability distribution,


Now E(Y²) = 14/5 and 2 E(X) = 2. 7/5 = 14/5
Hence, E(Y²) = 2E(X).

Q.31. A factory produces bulbs. The probability that any one bulb is defective is 1/50 and they are packed in boxes of 10. From a single box, find the probability that
(i) None of the bulbs is defective
(ii) Exactly two bulbs are defective
(iii) More than 8 bulbs work properly
Ans. Let X be the random variable denoting a bulb to be defective.
Here, n  = 10,
We know that P(X = r) =
(i) None of the bulbs is defective, i.e., r = 0

(ii) Exactly two bulbs are defective

(iii) More than 8 bulbs work properly We can say that less than 2 bulbs are defective
P(x < 2) = P(x = 0) + P(x = 1)



Q.32. Suppose you have two coins which appear identical in your pocket. You know that one is fair and one is 2-headed. If you take one out, toss it and get a head, what is the probability that it was a fair coin?
Ans. 
Let E₁ = Event that the coin is fair
E₂ = Event that the coin is 2-headed
and H = Event that the tossed coin gets head.

∴ Using Bayes’ Theorem, we get


Hence the required probability is 1/3.

Q.33. Suppose that 6% of the people with blood group O are left handed and 10% of those with other blood groups are left handed 30% of the people have blood group O. If a left handed person is selected at random, what is the probability that he/she will have blood group O?
Ans. Let E₁ = The event that a person selected is of blood group O
E₂ = The event that the person selected is of other group
and H = The event that selected person is left handed
∴ P(E₁) = 0.30   and P(E₂) = 0.70
P (HE₁) = 0.06 P (HE₂) = 0.10
So, from Bayes’ Theorem


Hence, the required probability is 9/44.

Q.34. Two natural numbers r, s are drawn one at a time, without replacement from the set S= {1 , 2 , 3 , . . . . , n } . Find P [ r ≤ p|s ≤ p ] , where p ∈ S.
Ans.Given that: S = {1, 2, 3, ..., n}
∴
Hence, the required probability is

Q.35. Find the probability distribution of the maximum of the two scores obtained when a die is thrown twice. Determine also the mean of the distribution.
Ans. Let X be the random variable scores when a die is thrown twice.
X = 1, 2, 3, 4, 5, 6 and S = {(1, 1), (1, 2), (2, 1), (2, 2), (1, 3), (2, 3), (3, 1), (3, 2),  (3, 3), (3, 4), (3, 5), ..., (6, 6)}
So,

Similarly
So, the required distribution is

Now, the mean E(X) =

Hence, the required mean = 161/36.

Q.36. The random variable X can take only the values 0, 1, 2. Given that P(X = 0) = P (X = 1) = p and that E(X²) = E[X], find the value of p.
Ans. Given that: X = 0, 1, 2
and P(X) at X = 0 and 1 is p. Let P(X) at X = 2 is x
⇒ p + p + x = 1
⇒ x = 1 – 2p
Now we have the following distributions.

∴ E(X) = 0.p + 1.p + 2(1 – 2p) = p + 2 – 4p = 2 – 3p
and E(X²) = 0.p + 1.p + 4(1 – 2p) = p + 4 – 8p = 4 – 7p
Given that: E(X²) = E(X)
∴ 4 – 7p = 2 – 3p
⇒ 4p = 2
p = 1/2
Hence, the required value of p is 1/2.

Q.37. Find the variance of the distribution:

Ans. We know that:
Variance (X) = E(X²) – [E(X)]²




Hence, the required variance is 665/324.

Q.38. A and B throw a pair of dice alternately. A wins the game if he gets a total of 6 and B wins if she gets a total of 7. It A starts the game, find the probability of winning the game by A in third throw of the pair of dice.
Ans. Let A₁ be the event of getting a total of 6
= {(2, 4), (4, 2), (1, 5), (5, 1), (3, 3)}
and B₁ be the event of getting a total of 7
= {(2, 5), (5, 2), (1, 6), (6, 1), (3, 4), (4, 3)}
Let P(A₁) is the probability, if A wins in a throw = 5/36
and P(B₁) is the probability, if B wins in a throw = 1/6
∴ The required probability of winning A in his third throw


Q.39. Two dice are tossed. Find whether the following two events A and B are independent:

where (x, y) denotes a typical sample point.
Ans. Given that:
A = {(x, y) : x + y = 11} and B = {(x, y) : x ≠ 5}
∴ A = {(5, 6), (6, 5)},
B = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
⇒ n(A) = 2, n(B) = 30 and n(A ∩ B) = 1
∴ P(A) = 2/36 = 1/18 and P(B) = =30/36 = 5/6
⇒ P(A). P(B) = 1/18.5/6 = 5/108 and P(A ∩ B) = 1/36
Since P(A). P(B) ≠ P(A ∩ B)
Hence, A and B are not independent.

Q.40. An urn contains m white and n black balls. A ball is drawn at random and is put back into the urn along with k additional balls of the same colour as that of the ball drawn. A ball is again drawn at random. Show that the probability of drawing a white ball now does not depend on k.
Ans. Let A be the event having m white and n black balls
E₁ = {first ball drawn of white colour}
E₂ = {first ball drawn of black colour}
E₃ = {second ball drawn of white colour}
∴

Now P(E₃) = P(E₁) . P (E_3/E₁) + P(E₂) . P (E_3/E₂)

Hence, the probability of drawing a white ball does not depend upon k.