Probability problems
- 1. One hundred identical coins each with probability p of showing up head are tossed. If 0 < p < 1 and the probability of heads showing on 50 coins is equals to that of heads on 51 coins. Then the value of p, is?
Options- A. 1/2
- B. 49/101
- C. 50/101
- D. 51/101 Discuss
Correct Answer: 51/101
Explanation:
Let N be the number of coins showing heads.
? Probability when N equal to 50 = Probability when N equal to 51
? 100C50 x P 50 x (1 - p)50 = 100C51 x (1 - p)49
? p = 51/101
- 2. Four whole number taken at random are multiplied together. The chance that the last digit in the product is 1, 3, 7, or 9 is?
Options- A. 16/625
- B. 1/210
- C. 8/125
- D. 4/25 Discuss
Correct Answer: 16/625
Explanation:
If the product of the four numbers ends in one of the digits 1, 3, 7 or 9, each number should have the last digit as one of these 4 digits.
? The number of favourable cases = 44
Total number of all possible cases = 104
Hence, the required probability =44/104
= 24/54 = 16/625
- 3. What is the probability that a card drawn at random from a pack of 52 cards is either a king or a spade?
Options- A. 17/52
- B. 4/13
- C. 3/13
- D. 13/52 Discuss
Correct Answer: 4/13
Explanation:
Required probability = 3/52 + 13/52 = 16/52 =4/13
[Hint Why 13/52 because there are 13 spades and why 3/52 instead of 4/52 (there are four kings) because one king is already counted in spades.]
- 4. A card is drawn from a well-shuffled pack of cards. The probability of getting a queen of club or a king of heart is?
Options- A. 1/52
- B. 1/26
- C. 1/13
- D. 1/39 Discuss
Correct Answer: 1/26
Explanation:
Total ways = 52
There is one queen of club and one king of heart.
? Favorable ways = 1 + 1 = 2
? Required probability = 2/52 = 1/26
- 5. If three unbiased coins are tossed simultaneously, then the probability of exactly two heads, is
Options- A. 1/8
- B. 2/8
- C. 3/8
- D. 4/8 Discuss
Correct Answer: 3/8
Explanation:
n(S) = 23 = 8
Let E = Event of getting exactly two heads
= {(H, H, T), (H, T ,H), (T, H, H)}
? n(E) = 3
? required probability = 3/8
- 6. Let E be the set of all integers with 1 at their unit places. The probability that a number chosen from {2, 3, 4, ............, 50} is an element of E, is?
Options- A. 5/49
- B. 4/49
- C. 3/49
- D. 2/49 Discuss
Correct Answer: 4/49
Explanation:
n(S) = 49
Favourable numbers are 11, 21, 31, 41.
? Required probability = 4/49
- 7. A number is selected at random from the set {1, 2, 3, ........, 50}. The probability that it is a prime, is
Options- A. 0.1
- B. 0.2
- C. 0.3
- D. 0.3 Discuss
Correct Answer: 0.3
Explanation:
n(s) = 50
Prime numbers are = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47
? n(E) = 15
? p(E) = 15/50 = 3/10 = 0.3
- 8. A committee of 3 members is to be selected to be selected out of 3 man and 2 women . What is the probability that the committee has at least one women?
Options- A. 1/10
- B. 9/20
- C. 9/10
- D. 1/20 Discuss
Correct Answer: 9/10
Explanation:
Required probability = (2C1 x 3C2 + 2C2 x 3C1) / (5C3) = 9/10
- 9. An urn contains 3 red and 4 green marbles. If three marbles are picked at random, what is the probability that two are green and one is red?
Options- A. 3/7
- B. 18/35
- C. 5/14
- D. 4/21 Discuss
Correct Answer: 18/35
Explanation:
Number of ways to select 3 marbles out of 7 marbles = n(s) = 7C3 = 35
Probability that 2 are green and 1 is red = n(E) = 4C2 x 3C1 = 18
? Required probability = 18/35
- 10. If two marbles are picked at random, what is the probability that both marbles are red?
Options- A. 1/6
- B. 1/3
- C. 2/15
- D. None of the above Discuss
Correct Answer: None of the above
Explanation:
Total number of marbles = 6 + 4 + 2 + 3 = 15
Ways of selection of two red marbles = n(E) = 6C2
Ways of selection of two marbles = n(S) = 15C2
So, required probability = (6C2) / (15C2)
= (6 x 5) / (15 x 14) = 1/7
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