2170 + Interview Questions in PROBABILITY IN GENERAL AWARENESS Page 44 InterviewSolution

2151.	A bag contains 12 pairs of socks. Four socks are picked up at random. Find the probability that there is at least one pair.
Answer» Correct Answer - `1-(.^(12)C_(4)xx 2^(4))/(.^(24)C_(4))` The number of ways of selecting 4 socks from 24 socks is `n(S) = .^(24)C_(4)`. The number of ways of selecting 4 socks from different pairs is `n(E) = .^(12)C_(4)xx2^(4)` `implies P(E) = (.^(12)C_(4) xx 2^(4))/(.^(24)C_(4))` Hence, the probability of getting at least one pair is `1 -(.^(12)C_(4) cc 2^(4))/(.^(24)C_(4))`

Discussion

2152.	Two coins are tossed simultaneously 500 times and the outcomes are noted as given below:Outcome:Two heads (HH)One head (HT or TH)No head (TT)Frequency:105275120If same pair of coins is tossed at random, find the probability of getting:(i) Two heads(ii) One head(iii) No head.
Answer» Given number of trials = 500 From the given table it is clear that, Number of outcomes of two heads (HH) = 105 Number of outcomes of one head (HT or TH) = 275 Number of outcomes of no head (TT) = 120 (i) Probability of getting two heads = frequency of getting 2 heads/ total number of trials = 105/500 = 21/100 (ii) Probability of getting one head = frequency of getting 1 heads/ total number of trials = 275/500 = 11/20 (iii) Probability of getting no head = frequency of getting no heads/ total number of trials = 120/500 = 6/25

2152.

Two coins are tossed simultaneously 500 times and the outcomes are noted as given below:Outcome:Two heads (HH)One head (HT or TH)No head (TT)Frequency:105275120If same pair of coins is tossed at random, find the probability of getting:(i) Two heads(ii) One head(iii) No head.

Answer»

Given number of trials = 500

From the given table it is clear that,

Number of outcomes of two heads (HH) = 105

Number of outcomes of one head (HT or TH) = 275

Number of outcomes of no head (TT) = 120

(i) Probability of getting two heads = frequency of getting 2 heads/ total number of trials

= 105/500

= 21/100

(ii) Probability of getting one head = frequency of getting 1 heads/ total number of trials

= 275/500

= 11/20

(iii) Probability of getting no head = frequency of getting no heads/ total number of trials

= 120/500

= 6/25

Discussion

2153.	A box contains two pair of socks of two colours (black and white). I have picked out a white sock. I pick out one more with my eyes closed. What is the probability that I will make a pair?
Answer» Given number of socks in the box = 4 Let B and W denote black and white socks respectively. Then we have S = {B, B, W, W} If a white sock is picked out, then the total no. of socks left in the box = 3 Number of white socks left = 2 – 1 = 1 Probability of getting white socks = number of white socks left in the box/ total number of socks left in the box = 1/3

2153.

A box contains two pair of socks of two colours (black and white). I have picked out a white sock. I pick out one more with my eyes closed. What is the probability that I will make a pair?

Answer»

Given number of socks in the box = 4

Let B and W denote black and white socks respectively. Then we have

S = {B, B, W, W}

If a white sock is picked out, then the total no. of socks left in the box = 3

Number of white socks left = 2 – 1 = 1

Probability of getting white socks = number of white socks left in the box/ total number of socks left in the box

= 1/3

Discussion

2154.	State True or False for the statements:Another name for the mean of a probability distribution is expected value.
Answer» TRUE Mean gives the average of values and if it is related with probability or random variable it is often called expected value.

2154.

State True or False for the statements:Another name for the mean of a probability distribution is expected value.

Answer»

TRUE

Mean gives the average of values and if it is related with probability or random variable it is often called expected value.

Discussion

2155.	Let X be a discrete random variable assuming values x1, x2, ..., xn with probabilities p1, p2, ..., pn, respectively. Then variance of X is given by(A) E (X2) (B) E (X2) + E (X) (C) E (X2) – [E (X)]2(D) √(E(X2 ) – [E (X)]2)
Answer» The correct answer is (C).

Discussion

2156.	If A and B are independent, then P (exactly one of A,B occurs)
Answer» Correct Answer - 1

Discussion

2157.	If A and B are two independent events, then P( A and B)=P(A)`cdot`P(B)
Answer» Correct Answer - 1

Discussion

2158.	State whether the statement is True or False.Three events A, B and C are said to be independent if P (A∩B∩C) = P (A) P (B) P (C).
Answer» False. Reason is that A, B, C will be independent if they are pairwise independent and P (A∩B∩C) = P (A) P (B) P (C).

Discussion

2159.	State True or False.Another name for the mean of a probability distribution is expected value.
Answer» Answer is True

Discussion

2160.	State True or False for the statements:If A and B are two independent events then P(A and B) = P(A).P(B)
Answer» TRUE If A and B are independent events. It implies- P(A ∩ B) = P(A)P(B) Thus, from the definition of independent event we say that statement is true.

2160.

State True or False for the statements:If A and B are two independent events then P(A and B) = P(A).P(B)

Answer»

TRUE

If A and B are independent events. It implies-

P(A ∩ B) = P(A)P(B)

Thus, from the definition of independent event we say that statement is true.

Discussion

2161.	State whether the statement is True or False. Let A and B be two independent events. Then P (A∩B) = P (A) + P (B)
Answer» False, because P (A∩B) = P (A) . P(B) when events A and B are independent.

Discussion

2162.	If A and B are independent events such that P (A) = p, P (B) = 2p and P (Exactly one of A, B) = 5/9 , then p = __________
Answer» p = 1/3, 5/12 [(1–p)(2p) + p(1– 2p) = 3p – 4p² = 5/9]

Discussion

2163.	State True or False for the statement:Two independent events are always mutually exclusive.
Answer» False If A and B are independent events. It implies- P(A ∩ B) = P(A)P(B) Through the above equation we can’t prove in any way that P(A∪B) = P(A) + P(B) It is only possible if either P(A) or P(B) = 0,which is not given in question. So, it is a false statement.

2163.

State True or False for the statement:Two independent events are always mutually exclusive.

Answer»

False

If A and B are independent events. It implies-

P(A ∩ B) = P(A)P(B)

Through the above equation we can’t prove in any way that

P(A∪B) = P(A) + P(B)

It is only possible if either P(A) or P(B) = 0,which is not given in question.

So, it is a false statement.

Discussion

2164.	If A and B′ are independent events then P (A′∪B) = 1 – ________
Answer» P (A′∪B) = 1 – P (A∩B′) = 1 – P (A) P (B′) (since A and B′ are independent).

2164.

If A and B′ are independent events then P (A′∪B) = 1 – ________

Answer»

P (A′∪B) = 1 – P (A∩B′) = 1 – P (A) P (B′)

(since A and B′ are independent).

Discussion

2165.	State True or False. Two independent events are always mutually exclusive.
Answer» Answer is False

Discussion

2166.	State True or False. If A and B are mutually exclusive events, then they will be independent also.
Answer» Answer is False

Discussion

2167.	State True or False.If A and B are two independent events then P(A and B) = P(A).P(B).
Answer» Answer is True

Discussion

2168.	(a) How many two-digit positive integers are multiples of 3?(b) What is the probability that a randomly chosen two-digit positive integer is a multiple of 3?
Answer» (a) 2 digit positive integers which are multiples of 3 are 12, 15, 18, ... , 99. Thus, there are 30 such integers. (b) 2-digit positive integers are 10, 11, 12, ... , 99. Thus, there are 90 such numbers. Since out of these, 30 numbers are multiple of 3, therefore, the probability that a randomly chosen positive 2-digit integer is a multiple of 3, is 30/90 = 1/3.

2168.

(a) How many two-digit positive integers are multiples of 3?(b) What is the probability that a randomly chosen two-digit positive integer is a multiple of 3?

Answer»

(a) 2 digit positive integers which are multiples of 3 are 12, 15, 18, ... , 99. Thus, there are 30 such integers.

(b) 2-digit positive integers are 10, 11, 12, ... , 99. Thus, there are 90 such numbers. Since out of these, 30 numbers are multiple of 3, therefore, the probability that a randomly chosen positive 2-digit integer is a multiple of 3, is 30/90 = 1/3.

Discussion

2169.	A basket contains 10 fruits in which three are rotten.If two fruits are selected from the basket, what is the probability that both the fruits are not rotten? (A) `7/15` (B) `8/15` (C) `1/15` (D) `10/15`
Answer» Initially, there are 10 fruits in basket and 3 of them are rotten. `:.`Probability of selecting first fruit that is not rotten ` = 7/10` Now, there are 9 fruits remaining in the basket and three of them are rotten. `:.` Probability of selecting another fruit that is not rotten `= 6/9 ` `:.` Probability of selecting both fruits that are not rotten ` = 7/10**6/9 = 7/15`.

Discussion

2170.	Three coins are tossed together. Find the probability of getting:(i) exactly two heads(ii) at least two heads(iii) at least one head and one tail(iv) no tails
Answer» When 3 coins are tossed together, Total no. of possible outcomes = 8 {HHH, HHT, HTH, HTT, THH,THT, TTH, TTT} (i) Probability of an event = (No. of favorable outcomes)/(Total no. of possible outcomes) Let E ⟶ event of getting exactly two heads No. of favourable outcomes = 3 {HHT, HTH, THH} Total no. of possible outcomes = 8 P(E) = 3/8 (ii) E ⟶ getting at least 2 Heads No. of favourable outcomes = 4 {HHH, HHT, HTH, THH} Total no. of possible outcomes = 8 P(E) = 4/8 = 1/2 (iii) E⟶ getting at least one Head & one Tail No. of favourable outcomes = 6 {HHT, HTH, HTT, THH, THT, TTH} Total no. of possible outcomes = 8 P(E) = 6/8 = 3/4 (iv) E ⟶ getting no tails No. of favourable outcomes = 1 {HHH} Total no. of possible outcomes = 8 P(E) = 1/8

2170.

Three coins are tossed together. Find the probability of getting:(i) exactly two heads(ii) at least two heads(iii) at least one head and one tail(iv) no tails

Answer»

When 3 coins are tossed together,

Total no. of possible outcomes = 8 {HHH, HHT, HTH, HTT, THH,THT, TTH, TTT}

(i) Probability of an event = (No. of favorable outcomes)/(Total no. of possible outcomes)

Let E ⟶ event of getting exactly two heads

No. of favourable outcomes = 3 {HHT, HTH, THH}

Total no. of possible outcomes = 8

P(E) = 3/8

(ii) E ⟶ getting at least 2 Heads

No. of favourable outcomes = 4 {HHH, HHT, HTH, THH}

Total no. of possible outcomes = 8

P(E) = 4/8 = 1/2

(iii) E⟶ getting at least one Head & one Tail

No. of favourable outcomes = 6 {HHT, HTH, HTT, THH, THT, TTH}

Total no. of possible outcomes = 8

P(E) = 6/8 = 3/4

(iv) E ⟶ getting no tails

No. of favourable outcomes = 1 {HHH}

Total no. of possible outcomes = 8

P(E) = 1/8

Discussion

Explore topic-wise InterviewSolutions in .

A bag contains 12 pairs of socks. Four socks are picked up at random. Find the probability that there is at least one pair.

Two coins are tossed simultaneously 500 times and the outcomes are noted as given below:Outcome:Two heads (HH)One head (HT or TH)No head (TT)Frequency:105275120If same pair of coins is tossed at random, find the probability of getting:(i) Two heads(ii) One head(iii) No head.

A box contains two pair of socks of two colours (black and white). I have picked out a white sock. I pick out one more with my eyes closed. What is the probability that I will make a pair?

State True or False for the statements:Another name for the mean of a probability distribution is expected value.

Let X be a discrete random variable assuming values x1, x2, ..., xn with probabilities p1, p2, ..., pn, respectively. Then variance of X is given by(A) E (X2) (B) E (X2) + E (X) (C) E (X2) – [E (X)]2(D) √(E(X2 ) – [E (X)]2)

If A and B are independent, then P (exactly one of A,B occurs)

If A and B are two independent events, then P( A and B)=P(A)`cdot`P(B)

State whether the statement is True or False.Three events A, B and C are said to be independent if P (A∩B∩C) = P (A) P (B) P (C).

State True or False.Another name for the mean of a probability distribution is expected value.

State True or False for the statements:If A and B are two independent events then P(A and B) = P(A).P(B)

State whether the statement is True or False. Let A and B be two independent events. Then P (A∩B) = P (A) + P (B)

If A and B are independent events such that P (A) = p, P (B) = 2p and P (Exactly one of A, B) = 5/9 , then p = __________

State True or False for the statement:Two independent events are always mutually exclusive.

If A and B′ are independent events then P (A′∪B) = 1 – ________

State True or False. Two independent events are always mutually exclusive.

State True or False. If A and B are mutually exclusive events, then they will be independent also.

State True or False.If A and B are two independent events then P(A and B) = P(A).P(B).

(a) How many two-digit positive integers are multiples of 3?(b) What is the probability that a randomly chosen two-digit positive integer is a multiple of 3?

A basket contains 10 fruits in which three are rotten.If two fruits are selected from the basket, what is the probability that both the fruits are not rotten? (A) `7/15` (B) `8/15` (C) `1/15` (D) `10/15`

Three coins are tossed together. Find the probability of getting:(i) exactly two heads(ii) at least two heads(iii) at least one head and one tail(iv) no tails