Questions: Probability

Q1. Two events have probabilities 0.6 and 0.3 for A and B, and the overlap has probability 0.2. Find both conditional probabilities.

Answer (English): The conditional probability of A given B is the overlap divided by the probability of B, which is 0.2/0.3 = 2/3. The conditional probability of B given A is the overlap divided by the probability of A, which is 0.2/0.6 = 1/3. The two answers differ because the two conditions reduce the sample space differently.

Hinglish: Overlap ko condition wale event se divide karo: P(A|B) = 0.2/0.3 = 2/3 aur P(B|A) = 0.2/0.6 = 1/3. Dono jawab alag hain kyunki condition alag hai; bar ke baad wala event dhyan se chuno.

Q2. In a family with two children, find the probability that both are boys given that at least one child is a boy.

Answer (English): The four equally likely cases are boy-boy, boy-girl, girl-boy and girl-girl. The condition that at least one is a boy leaves three of them, and only boy-boy has both boys. So the required conditional probability is one third, not one half, because learning that at least one is a boy removes only the girl-girl case.

Hinglish: Chaar cases hain: BB, BG, GB, GG. "At least one boy" ke baad teen cases bachte hain, aur unmein sirf BB dono ladke hain, isliye answer 1/3 hai, 1/2 nahi; condition sirf GG ko hataati hai.

Q3. Two dice are thrown and the sum is found to be 6. Find the conditional probability that 4 appeared on at least one die.

Answer (English): The outcomes with sum 6 are (1,5), (2,4), (3,3), (4,2) and (5,1), so the condition has five equally likely cases. Of these, two contain a 4, namely (2,4) and (4,2). The conditional probability is therefore 2/5.

Hinglish: Sum 6 ke outcomes hain (1,5), (2,4), (3,3), (4,2), (5,1), yaani paanch equally likely cases. Inmein 4 wale do hain: (2,4) aur (4,2), isliye 2/5.

Q4. Two events A and B are independent with probabilities 0.4 and 0.5. Find the probability of both, the probability of at least one, and the conditional probability of A given B.

Answer (English): Independence makes the probability of both equal to the product, 0.2. The addition rule then gives the probability of at least one as 0.4 + 0.5 - 0.2 = 0.7. The conditional probability of A given B is the overlap divided by the probability of B, 0.2/0.5 = 0.4, which equals the probability of A, exactly as independence requires.

Hinglish: Independence se dono ki probability 0.2, addition rule se at least one ki 0.7, aur P(A|B) = 0.2/0.5 = 0.4, jo P(A) ke barabar hai, jaise independence mein hona chahiye.

Q5. Bag A has 3 red and 4 white balls while bag B has 5 red and 6 white balls. A bag is chosen at random and one ball is drawn. Find the probability that it is red.

Answer (English): The two bags are the cases, each chosen with probability one half. The probability of red from the first bag is 3/7 and from the second is 5/11. By the theorem of total probability the answer is half of 3/7 plus half of 5/11, which simplifies to 34/77.

Hinglish: Dono bags ki probability 1/2 hai. First se red 3/7 aur second se red 5/11, isliye total probability se answer = (1/2)(3/7) + (1/2)(5/11) = 34/77.

Q6. Two factories produce 60 per cent and 40 per cent of the items with defect rates 2 per cent and 3 per cent. A defective item is found. Find the probability that it came from the first factory.

Answer (English): By Bayes' theorem the numerator for the first factory is 0.6 times 0.02, which is 0.012. The total probability of a defective item is 0.012 plus 0.4 times 0.03, also 0.012, so the denominator is 0.024. The posterior probability is 0.012/0.024 = 1/2. The two factories are equally likely to have produced the defective item.

Hinglish: Bayes' theorem se numerator = 0.6 × 0.02 = 0.012, aur total = 0.012 + 0.4 × 0.03 = 0.024. Posterior = 0.012/0.024 = 1/2, yaani dono factories barabar chance wali hain.

Q7. A random variable has probabilities k, 2k and 3k for the values 0, 1 and 2. Find k, then the mean and variance.

Answer (English): The probabilities add to 1, so 6k = 1 and k = 1/6. The distribution is 1/6, 1/3 and 1/2 for 0, 1 and 2. The mean is 0 + 1/3 + 1 = 4/3. The mean of the squares is 0 + 1/3 + 2 = 7/3, so the variance is 7/3 - 16/9 = 5/9.

Hinglish: Sum 1 se k = 1/6, distribution 1/6, 1/3, 1/2. Mean = 0 + 1/3 + 1 = 4/3, E(X²) = 7/3, isliye variance = 7/3 − 16/9 = 5/9.

Q8. A fair coin is tossed five times. Find the probability of exactly three heads, and the mean and variance of the number of heads.

Answer (English): The count of heads is binomial with five trials and success probability one half. The probability of exactly three heads is the binomial coefficient five choose three times one half to the fifth power, which is 10/32 = 5/16. The mean is five times one half, 2.5, and the variance is five times one half times one half, 1.25.

Hinglish: Heads ki count binomial distribution follow karti hai: n = 5, p = 1/2. Exactly 3 heads ki probability 5/16, aur mean = 2.5, variance = 1.25.

Q9. The probability that a bulb is defective is 0.1. In a lot of 20 bulbs, find the probability that at most one is defective.

Answer (English): This is binomial with n = 20, p = 0.1 and q = 0.9. At most one defective means zero or one, so the answer is the probability of zero plus the probability of one. That is 0.9 to the power 20 plus 20 times 0.1 times 0.9 to the power 19, which simplifies to 2.9 times 0.9 to the power 19, about 0.392.

Hinglish: n = 20, p = 0.1, q = 0.9. At most 1 ka matlab 0 ya 1 defective, isliye answer = P(0) + P(1) = 2.9 × (0.9)^19 ≈ 0.392.

Q10. Two cards are drawn one after the other without replacement from a pack of 52. Find the probability that the first card is red and the second is a jack.

Answer (English): Split the first draw into a red jack and a red card that is not a jack. If the first card is a red jack, the second draw has 3 jacks among 51 cards; if the first is a red non-jack, the second draw has 4 jacks among 51. The total is two-fifty-seconds times three-fifty-firsts plus twenty-four-fifty-seconds times four-fifty-firsts, which is 102/2652 = 1/26.

Hinglish: Pehle draw ko red jack aur red non-jack mein todo. Pehla red jack ho to 51 mein 3 jack bache, aur red non-jack ho to 51 mein 4 jack. Isliye total = (2/52)(3/51) + (24/52)(4/51) = 102/2652 = 1/26.

To check how much of this stuck, take the five-question quiz below. Each answer explains the reasoning, so read the explanation even when you get it right.

Quiz: the five question check for this chapter is set from the same topics. Attempt it after the questions above, then revisit any topic page where you lost a mark.