Question 1

A random variable can assume the values 0, 1, 2,..., n, and its corresponding frequencies are proportional to the coefficients nC0, nC1, nC2,..., nCn. What is the variance of this distribution?

Accepted Answer

n/4. Frequencies proportional to nC0,nC1,...,nCn give a binomial distribution with p=1/2. Variance = n·p·(1-p) = n·(1/2)·(1/2) = n/4.

Question 2

From the integers 1, 2, 3,..., 2004, two distinct numbers x and y are selected at random without replacement. What is the probability that x³ + y³ is divisible by 3?

Accepted Answer

1/3. Since n^3 = n (mod 3), x^3+y^3 is divisible by 3 iff x+y is. Among 1..2004 there are 668 numbers in each residue class 0,1,2. Favourable unordered pairs = C(668,2) (both =0) + 668*668 (one =1, one =2). Dividing by C(2004,2) gives exactly 1/3.

Question 3

One card is selected at random from a standard deck of 52 cards. A player wagers that the card will be a spade or an ace. What are the odds against the player winning the wager?

Accepted Answer

9: 4. The player wins if they draw a spade (13 cards) or an ace (4 cards), but since one of the aces is also a spade, there are 16 winning cards in total. Therefore, the odds against winning are calculated by comparing the number of losing cards (36) to the winning cards (16), simplifying to 9 losing cards for every 4 winning cards.

Question 4

When n objects are assigned randomly to n people, what is the probability that at least one person receives no object?

Accepted Answer

1 - (n-1)!/(n^(n-1)). The correct option represents the probability that at least one person does not receive an object, which is derived from the total arrangements of objects and the arrangements where each person receives at least one object. By calculating the complement of the scenario where everyone receives at least one object, we arrive at the expression 1 - (n-1)!/(n^(n-1)).

Question 5

For two arbitrary events M and N, what is the probability that one and only one of them happens?

Accepted Answer

P(M)+P(N)-2 P(M∩N). 'One and only one happens' means (M and not N) or (N and not M). P(M only)=P(M)-P(M∩N) and P(N only)=P(N)-P(M∩N); adding gives P(M)+P(N)-2P(M∩N).

Question 6

A four-digit number is made using the digits 1, 2, 3, and 4 without repeating any digit. What is the probability that the resulting number is odd?

Accepted Answer

None of these. Among the four digits two are odd (1,3) and two even (2,4). The last digit is equally likely to be any of the four, so P(odd) = 2/4 = 1/2, which is not listed; the answer is None of these.

Question 7

Out of 600 bolts, 20% are oversized and 10% are undersized. The rest are acceptable. If one bolt is chosen at random, what is the probability that it is acceptable?

Accepted Answer

7/10. To find the probability of selecting an acceptable bolt, first calculate the number of oversized and undersized bolts: 20% of 600 is 120 oversized, and 10% is 60 undersized, totaling 180 non-acceptable bolts. Therefore, the number of acceptable bolts is 600 - 180 = 420, leading to a probability of 420/600, which simplifies to 7/10.

Question 8

In a leap year, what is the probability that the year contains 53 Fridays or 53 Saturdays?

Accepted Answer

3/7. A leap year has 366 = 52 weeks + 2 extra consecutive days. P(53 Fri) = 2/7, P(53 Sat) = 2/7, and P(both, i.e. extra days are Fri&Sat) = 1/7. So P = 2/7 + 2/7 - 1/7 = 3/7.

Question 9

Using the digits 1, 2, 3, 4 and 5, a two-digit number is formed. What is the probability that this number is divisible by 4?

Accepted Answer

1/5. With digits 1-5, two-digit numbers divisible by 4 are 12, 24, 32, 52 (and 44 if repetition allowed). Without repetition: 4 favourable out of 5*4 = 20, probability 4/20 = 1/5. With repetition: 5 out of 25 = 1/5. Either way the probability is 1/5.

Question 10

Two values p and q are independently selected at random from the set {1, 2, 3, 4}, with replacement. What is the probability that p² is at least 4q?

Accepted Answer

7/16. From {1,2,3,4} with replacement: q=1 needs p>=2 (3 ways), q=2 needs p>=3 (2), q=3 needs p>=4 (1), q=4 needs p=4 (1). Total 7 favourable out of 16, so probability = 7/16.

Question 11

Two integers m and n are selected independently and uniformly from the set {1, 2,..., 100}. What is the probability that the number 7^m + 7ⁿ is divisible by 5?

Accepted Answer

1/4. 7^k mod 5 cycles through 2,4,3,1. The sum 7^m + 7^n is divisible by 5 only for residue pairs (2,3),(3,2),(4,1),(1,4). For each m, exactly 1 of 4 residue classes of n works, giving probability 1/4.

Question 12

Let A and B be events with P(A)=1/2 and P(B)=2/3. Which statement below is false?

Accepted Answer

P(A ∩ B') is at least 1/3. The statement P(A ∩ B') is at least 1/3 is false because it suggests that the probability of event A occurring while event B does not occur is greater than or equal to 1/3, which is not supported by the given probabilities of A and B.

Question 13

Let A, B, and C be three events. The probability that exactly one of A and B occurs is 1, that exactly one of B and C occurs is 1/2, and that exactly one of C and A occurs is 1/4. If the probability that all three events occur together is a², then the probability that at least one of A, B,

Accepted Answer

> 1/2. Adding the three 'exactly one' probabilities: [P(A)+P(B)-2P(AB)]+... = 1 + 1/2 + 1/4 = 7/4, so (sum of singles) - (sum of pairwise) = 7/8. Then P(at least one) = (sum singles) - (sum pairs) + P(ABC) = 7/8 + a^2 >= 7/8 > 1/2.

Question 14

If the probabilities of two events A and B are equal, with P(A)=P(B)=x, and P(A ∩ B)=P(A' ∩ B')=1/3, what is the value of x?

Accepted Answer

1/2. P(A'/\B') = 1 - P(AUB) = 1/3, so P(AUB) = 2/3. Then P(AUB) = P(A)+P(B)-P(A/\B) = 2x - 1/3 = 2/3, giving 2x = 1, x = 1/2.

Question 15

A container holds 13 red, 14 green, and 15 white balls. Let p1 denote the probability that exactly 2 white balls are obtained when 4 balls are selected. Now the number of balls of each colour is doubled. Let p2 denote the probability that exactly 4 white balls are obtained when 8 balls are

Accepted Answer

p1 > p2. p1 = C(15,2)*C(27,2)/C(42,4) ≈ 0.329. After doubling: p2 = C(30,4)*C(54,4)/C(84,8) ≈ 0.199. Since 0.329 > 0.199, we have p1 > p2.

Question 16

Let A and B be events with P(A ∪ B)=3/4, P(A ∩ B)=1/4, and P(Ā)=2/3. Find the value of P(Ā ∩ B).

Accepted Answer

5/12. P(A)=1-2/3=1/3. From P(AuB)=P(A)+P(B)-P(A∩B): 3/4=1/3+P(B)-1/4, so P(B)=2/3. Then P(A'∩B)=P(B)-P(A∩B)=2/3-1/4=5/12.

Question 17

The probabilities of three mutually exclusive events are given by (1+4p)/4, (1-p)/2, and (1-2p)/2. What is the value of p?

Accepted Answer

1/2. For mutually exclusive events the probabilities must each lie in [0,1] and sum to at most 1. Testing p=1/2: probs are 3/4, 1/4, 0, summing to exactly 1 and all valid. p=1/3 gives a sum of 13/12 > 1, which is impossible. So p = 1/2.

Question 18

From the integers 1, 2, 3,..., 2004, two distinct numbers x and y are selected at random without replacement. The probability that x³ + y³ is divisible by 3 equals

Accepted Answer

1/3. The sum of cubes, x³ + y³, is divisible by 3 if both x and y have the same remainder when divided by 3. Since the integers 1 to 2004 yield three possible remainders (0, 1, 2), the probability of selecting two numbers with the same remainder is calculated by considering the combinations of pairs from each remainder class, leading to a final probability of 1/3.

Question 19

Let A, B and C be events with P(A) = 0.3, P(B) = 0.4, P(C) = 0.8, P(A ∩ B) = 0.08, P(A ∩ C) = 0.28 and P(A ∩ B ∩ C) = 0.09. If P(A ∪ B ∪ C) is at least 0.75, then

Accepted Answer

0.23 ≤ P(B ∩ C) ≤ 0.48. P(AUBUC)=PA+PB+PC-P(AB)-P(AC)-P(BC)+P(ABC)=1.23-P(BC). Requiring this ≥0.75 gives P(BC)≤0.48, and requiring it ≤1 gives P(BC)≥0.23, so 0.23 ≤ P(B∩C) ≤ 0.48.

Question 20

Two numbers x and y are independently selected with replacement from the set {0, 1, 2, 3,..., 10}. What is the probability that the absolute difference between them exceeds 5?

Accepted Answer

30/121. Total ordered pairs = 11x11 = 121. For |x-y|=d (d=6..10) there are 2(11-d) ordered pairs: 10+8+6+4+2 = 30. Probability = 30/121.

Question 21

A fair coin is flipped a total of m + n times, where m is greater than n. The probability that there is a run of exactly m consecutive heads is

Accepted Answer

(n + 2)/2^(m+1). The probability of obtaining a run of m consecutive heads when a fair coin is tossed m+n times (m>n) is (n+2)/2^(m+1). Direct enumeration for small m,n (e.g. m=2,n=1 gives 3/8; m=3,n=2 gives 1/4) confirms this formula exactly.

Question 22

For two events A and B, suppose that P(A ∪ B) is at least 3/4 and the intersection probability satisfies 1/8 ≤ P(A ∩ B) ≤ 3/8.

Statement-1: P(A) + P(B) is at least 7/8.
Statement-2: P(A) + P(B) is at most 11/8.

Choose the correct option:
(a) Both statements are true, and Statement-2 corr

Accepted Answer

Both statements are true, but Statement-2 does not correctly explain Statement-1.. Since P(A)+P(B)=P(AuB)+P(A∩B): minimum = 3/4 + 1/8 = 7/8 (Statement 1 true), maximum = 1 + 3/8 = 11/8 (Statement 2 true). Both are true but they use different bounds, so Statement 2 does not explain Statement 1 -> option (b).

Question 23

Let A and B be two events with P(Ā)=0.3, P(B)=0.4, and P(A ∩ B̄)=0.5. Find P(B | A ∪ B̄).

Accepted Answer

0.25. P(A)=0.7; P(A∩B̄)=0.5 gives P(A∩B)=0.2. B∩(A∪B̄)=A∩B=0.2. P(A∪B̄)=0.7+0.6-0.5=0.8. So P(B|A∪B̄)=0.2/0.8=0.25.

Question 24

A bag has 3 white, 2 black, and 4 red balls. Four balls are selected at random with replacement. What is the probability that exactly one of the four balls drawn is white?

Accepted Answer

32/81. P(white)=3/9=1/3 each draw (with replacement). P(exactly one white in 4) = C(4,1)(1/3)(2/3)^3 = 4*8/81 = 32/81.

Question 25

For each k = 1, 2, 3, the box Bₖ has k red balls and (k + 1) white balls. The probabilities of choosing B₁, B₂ and B₃ are 1/2, 1/3 and 1/6 respectively. One box is picked at random and then one ball is taken out. Given that the ball drawn is red, what is the probability that it was drawn f

Accepted Answer

14/39. The probability of drawing a red ball from box B₂, given that a red ball was drawn, is calculated using Bayes' theorem. By considering the total probability of drawing a red ball from all boxes and the specific probability from B₂, we find that the correct probability is 14/39.

Question 26

Six coins are tossed at the same time 64 times. If getting a head is taken as a success, what is the expected number of times that there will be at least 3 successes?

Accepted Answer

42. The expected number of times to get at least 3 heads when tossing 6 coins follows a binomial distribution. The probability of getting at least 3 heads in one toss is calculated, and multiplying this probability by the number of trials (64) yields the expected value, which is 42.

Question 27

Given that P(A)=2/5, P(B)=3/10, and P(A ∩ B̄)=1/5, find the value of P(A'|B') × P(B'|A').

Accepted Answer

25/42. P(A∩B)=P(A)-P(A∩B')=2/5-1/5=1/5, P(AuB)=2/5+3/10-1/5=1/2, so P(A'∩B')=1/2. Then P(A'|B')=(1/2)/(7/10)=5/7 and P(B'|A')=(1/2)/(3/5)=5/6; product = 25/42.

Question 28

A binomial random variable X has mean 6 and variance 2. The probability that 5 ≤ X ≤ 7 is

Accepted Answer

4672/6561. The mean and variance of a binomial random variable are related to the number of trials and the probability of success. Given the mean of 6 and variance of 2, we can determine the parameters of the binomial distribution and calculate the probability for the specified range, leading to the correct answer of 4672/6561.

Question 29

If n letters, each with its own correctly addressed envelope, are inserted into the envelopes in a completely random manner, what is the probability that all n letters end up in their proper envelopes?

Accepted Answer

None of these. All n letters in correct envelopes corresponds to a single arrangement (the identity), so the probability is 1/n!. This value is not among the listed series, so the answer is 'None of these'.

Question 30

A bag has n balls. The probability that exactly r of them are white is proportional to r² for 0 ≤ r ≤ n. If one ball is selected at random and it turns out to be white, then the probability that all the balls in the bag are white is:

Accepted Answer

4n/(n+1)². The correct option is right because the probability of selecting a white ball is influenced by the total number of white balls, which follows a quadratic relationship. Given that the probability of exactly r white balls is proportional to r², the calculations lead to the conclusion that the probability of all balls being white, given one is white, simplifies to 4n/(n+1)².

Question 31

A random variable X can take rational values of the types n/(n+1) and (n+1)/n, where n = 1, 2, 3,.... If P(X = n/(n+1)) = P(X = (n+1)/n) = (1/2)^(n+1), then which of the following statements is true?

Accepted Answer

All of the above are true. All the statements are true because they reflect the distribution of probabilities for the values of X. The probabilities assigned to each value ensure that the cumulative probabilities for the intervals specified in the statements maintain the relationships described, confirming the validity of each statement.

Question 32

In a multiple-choice test with four alternatives, a candidate responds in one of three ways: by guessing, by copying, or by actually knowing the answer. The probability that he guesses is 1/3, and the probability that he copies is 1/6. If he copies, the chance that his response is correct

Accepted Answer

24/29. P(know)=1-1/3-1/6=1/2. P(correct)=(1/3)(1/4)+(1/6)(1/8)+(1/2)(1)=1/12+1/48+1/2=29/48. By Bayes, P(know|correct)=(1/2)/(29/48)=24/29.

Question 33

Events E and F are independent. Given that P(E) = 0.3 and P(E ∪ F) = 0.5, find the value of P(E|F) - P(F|E).

Accepted Answer

1/70. P(EuF)=P(E)+P(F)-P(E)P(F)=0.5 => 0.3+0.7P(F)=0.5 => P(F)=2/7. With independence P(E|F)=P(E)=3/10 and P(F|E)=P(F)=2/7, so difference = 3/10 - 2/7 = 1/70.

Question 34

A man has a probability of 2/5 of hitting a target on a single shot. If he shoots at the target k times, where k is a given integer, what is the least value of k for which the probability of hitting the target at least once exceeds 7/10?

Accepted Answer

3. P(at least one hit) = 1 - (3/5)^k > 7/10 means (3/5)^k < 3/10. For k=2, (3/5)^2 = 0.36 > 0.3; for k=3, 0.216 < 0.3. So the least k is 3.

Question 35

If A and B are events with P(A) ≠ 0 and P(B) ≠ 1, then the value of P(Ā | B̄) is:

Accepted Answer

1 - P(A ∪ B) / P(B̄). P(A'|B') = P(A' ∩ B')/P(B') = P((A∪B)')/P(B') = (1 - P(A∪B))/P(B'). The option P(A')/P(B') can exceed 1 and is invalid, while 1 - P(A∪B)/P(B') (read as the standard NCERT form) gives the correct value.

Question 36

Let X be a binomial random variable with parameters n and p, where 0<p<1. If the ratio P(X=r) / P(X=n-r) does not depend on n or r, then the value of p is

Accepted Answer

1/2. P(X=r)/P(X=n-r) = (p/q)^r (q/p)^(n-r) = (p/q)^(2r-n). This is independent of n and r only if p/q=1, i.e. p=1/2.

Question 37

A student takes tests I, II and III. He is said to be successful if he clears either both tests I and II, or both tests I and III. The probabilities of passing tests I, II and III are p, q and 1/2 respectively. If the probability that the student is successful is 1/2, then the relation bet

Accepted Answer

pq + p = 1. With independence, P(success)=pq + p(1/2) - pq(1/2) = (p/2)(1+q). Setting this to 1/2 gives p(1+q)=1, i.e. pq + p = 1.

Question 38

A person moves one unit forward with probability 0.4 and one unit backward with probability 0.6. After 11 such moves, the probability that he is exactly one unit from his starting position is

Accepted Answer

462 × (6/25)⁵. To be one unit away after 11 moves: (F=6,B=5) or (F=5,B=6). Probability = C(11,6)(0.4^6 0.6^5 + 0.4^5 0.6^6) = 462*0.4^5*0.6^5 = 462*(6/25)^5.

Question 39

Let X and Y be independent random variables with distributions B(5, 1/2) and B(7, 1/2), respectively. What is the value of P(X + Y = 3)?

Accepted Answer

55/1024. Sum of independent binomials with the same p: X+Y~B(12,1/2). So P(X+Y=3)=C(12,3)/2^12=220/4096=55/1024.

Question 40

Let X be a binomial random variable with n = 8 and p = 1/2. What is the value of P(|X - 4| ≤ 2)?

Accepted Answer

119/128. P(|X-4|<=2)=P(2<=X<=6)=1-[P(0)+P(1)+P(7)+P(8)] = 1-(1+8+8+1)/256 = 1-18/256 = 238/256 = 119/128.

Question 41

A bag has 4 red balls and 4 blue balls. If 4 balls are drawn successively without replacement, what is the probability that the colours come out in alternating order?

Accepted Answer

6/35. Alternating draws are RBRB or BRBR. P(RBRB) = (4/8)(4/7)(3/6)(3/5) = 144/1680, and P(BRBR) is the same. Total = 288/1680 = 6/35.

Question 42

Let X be a Poisson random variable. If P(X = 1) equals P(X = 2), then the value of P(X = 4) is

Accepted Answer

2/(3e²). For a Poisson variable, P(X=1)=P(X=2) means e^-L L = e^-L L^2/2, so L=2. Then P(X=4)=e^-2 * 2^4/4! = 16/(24 e^2) = 2/(3 e^2).

Question 43

A company produces electric bulbs, and 3% of them are faulty. If a sample of 100 bulbs is considered and the Poisson approximation is used, what is the probability that the sample contains exactly one defective bulb?

Accepted Answer

.15. lambda = 100*0.03 = 3. P(X=1) = lambda e^(-lambda) = 3 e^(-3) approx 0.149, which is closest to 0.15.

Question 44

A discrete random variable X is distributed as follows:
X: 1 2 3 4 5 6 7 8
P(X): 0.15 0.23 0.12 0.10 0.20 0.08 0.07 0.05
If E denotes the event that X is a prime number and F denotes the event that X < 4, then the value of P(E ∪ F) is

Accepted Answer

0.77. Primes are {2,3,5,7}, X<4 is {1,2,3}, so E∪F={1,2,3,5,7}. P=0.15+0.23+0.12+0.20+0.07=0.77.

Question 45

In a five-match Test series between India and West Indies, suppose each match is independent and the probability that India wins any one match is 1/2. What is the probability that India’s second victory comes in the third Test?

Accepted Answer

1/4. The second win occurring in the third Test means exactly one win in the first two matches and a win in the third: C(2,1)(1/2)^2 * (1/2) = 2*(1/4)*(1/2) = 1/4.

Question 46

A factory produces items, each of which is defective with probability p. From a lot, a sample of n items is taken with replacement. If the sample contains no defective item, the lot is accepted; if it contains more than two defective items, the lot is rejected. If the sample contains exact

Accepted Answer

qⁿ + npq^(n-1) (q^m + mpq^(m-1)) + n(n-1)/2 p² q^(n+m-2). Accept if first sample has 0 defects (q^n); if exactly 1 defect (npq^(n-1)) then second sample needs <=1 defect (q^m+mpq^(m-1)); if exactly 2 defects (n(n-1)/2 p^2 q^(n-2)) then second sample needs 0 defects (q^m). Total = q^n + npq^(n-1)(q^m+mpq^(m-1)) + n(n-1)/2 p^2 q^(n+m-2).

Question 47

A machine produces items with defect probability 0.2. If 10 items from this machine are chosen at random, what is the probability that at most one of them is defective?

Accepted Answer

none of these. The probability of at most one defective item among 10 can be calculated using the binomial distribution formula, which shows that the resulting probability does not match any of the provided options, confirming that 'none of these' is the correct answer.

Question 48

Two independent witnesses, A and B, give the same statement. The probability that A tells the truth is x, and the probability that B tells the truth is y. If both agree on a statement, then the probability that the statement is true is

Accepted Answer

xy/(1 - x - y + 2xy). P(true and agree)=xy, P(false and agree)=(1-x)(1-y). P(true|agree)=xy/[xy+(1-x)(1-y)] = xy/(1 - x - y + 2xy).

Question 49

Statement 1: If a random variable X has a binomial distribution with mean a and variance b, then a²/(a-b) is a positive integer.
Statement 2: For a binomial distribution, the mean is greater than the variance.

Choose the correct option:
(a) Statement 1 is false and Statement 2 is true
(b)

Accepted Answer

(c). Statement 1 is true because for a binomial distribution, the mean (a) and variance (b) can be expressed in terms of the number of trials and probability, leading to the conclusion that a²/(a-b) is a positive integer. Statement 2 is also true as, in a binomial distribution, the mean is indeed greater than the variance when the probability of success is less than 0.5, but it does not explain why the first statement holds.

Question 50

A container holds 6 marbles that are red, 5 that are blue, and 4 that are white. If 4 marbles are drawn at random without replacement, what is the probability that the selection includes at least one marble of each colour?

Accepted Answer

48/91. With 4 marbles drawn from 6+5+4=15, the favourable colour counts are the (2,1,1) splits. Summing C(6,r)C(5,b)C(4,w) over all such splits gives 720 favourable out of C(15,4)=1365, i.e. 720/1365 = 48/91.

JEE Main Maths: Probability questions with solutions

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