Probability problems
- 1. What is the possibility of getting at least 6 heads, if eight coins are tossed simultaneously?
Options- A. 37/256
- B. 25/57
- C. 1/13
- D. None of these Discuss
Correct Answer: 37/256
Explanation:
Required probability = Probability of getting 2 tails + Probability of getting 1 tail + Probability of getting no tail.
= 8C2 x 1/256 + 8C1 x 1/256 + 8C0 x 1/256
= 37/256
- 2. A and B aim a target. The Probability that A hits the target is 5/7 and B hits the target in 7/10. What is the probability that exactly one of them hits the target?
Options- A. 21/10
- B. 29/70
- C. 31/70
- D. 33/70 Discuss
Correct Answer: 29/70
Explanation:
Required probability = P(A). P(B) + P (A ) . P(B)
= 5/7 . 3/10 + 2/7 . 7/10
= 29/70
- 3. A bag contains 5 white, 7 red and 8 black balls. If 4 balls are drawn one by one with replacement, what is the probability that all are white?
Options- A. 1/256
- B. 1/16
- C. 4/20
- D. 4/8 Discuss
Correct Answer: 1/256
Explanation:
Probability that first ball drawn is white = 5C1 = 1/4
Since, balls are drawn with replacement, hence all the four events will have equal probability.
Therefore, required probability = 1/4 x 1/4 x 1/4 x 1/4 = 1/256
- 4. There are 8 blue and 4 white balls in a bag . A ball is drawn at random. Without replacing it another ball is drawn. Find the probability that both the balls drawn are blue?
Options- A. 7/33
- B. 14/33
- C. 11/33
- D. 2/33 Discuss
Correct Answer: 14/33
Explanation:
Total number of balls in the bag = 12
Probability of drawing one blue ball in the first draw P(E1) = 8C1 / 12C1
= 8/12 = 2/3
After the first drawn of a blue ball, now there are 7 blue and 4 white ball in the bag. Total number of the balls in the bag is 11.
Probability of drawing one blue ball in the second drawn = P(E2) = 7C1 /11C1 = 7/11
? Probability that both are blue P(E1 ? E2) = P(E1) x P(E2)
= 2/3 x 7/11 = 14 / 33
- 5. Two dice are thrown simultaneously. What is the probability of getting a number other than 4 on any dice?
Options- A. 25/36
- B. 1/3
- C. 17/36
- D. 2/3 Discuss
Correct Answer: 25/36
Explanation:
Here, n(S) = 6 x 6 = 36 and E = Event of getting a number other than 4 on any dice
= {(1, 1), (1, 2), (1, 3), (1, 5), (1, 6), (2, 1), (2, 2) , (2, 3) , (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 5), (3, 6), (5, 1) , (5, 2) , (5, 3), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 5), (6, 6) }
? P(E) = n(E) / n(S) = 25/36
- 6. Number 1 to 25 are marked on tokens of equals size one on each. A token is drawn at random. Find the probability of getting a number divisible by both 2 and 3?
Options- A. 4/25
- B. 7/25
- C. 2/25
- D. 9/25 Discuss
Correct Answer: 4/25
Explanation:
Total number of ways of selecting one number from 25 number is 25C1 ways.
Let E be the event of drawing a number divisible by both 2 and 3
i.e., (if a number is divisible by both 2 and 3, it is divisible by 6 also)
E = {6, 12, 18, 24}
? Required probability = P(E) = n(E) / n(S) = 4/25
- 7. A complete cycle of a traffic light takes 60 s. During each cycle the light is green for 25 s, yellow for 5 s and red for 30 s. At a randomly chosen time, the probability that the light will not be green is?
Options- A. 1/3
- B. 1/4
- C. 5/12
- D. 7/12 Discuss
Correct Answer: 7/12
Explanation:
Time taken for complete cycle = 60 s
Probability that light will be green = (Time for which light is green)/(time taken for complete cycle)
= 25/60 = 5/12
? Probability that light will not be green = 1- 5/12 = 7/12
- 8. The probability that a man will live 10 more year is 1/4 and the probability that his wife will live 10 more year is 1/3. Then, the probability that none of them will be alive after 10 yr, is?
Options- A. 5/12
- B. 1/2
- C. 7/12
- D. 11/12 Discuss
Correct Answer: 1/2
Explanation:
Let the name of the man be A and that of his wife be B. Then ,P(A will not be alive after 10 yr and B will not be alive after 10yr) = 3/4 x 2/3 = 1/2
- 9. The letters B, G, I, N, R are rearranged to form the word 'BRING'. Find its probability?
Options- A. 1/120
- B. 1/54
- C. 1/24
- D. 5/5 x 42 Discuss
Correct Answer: 1/120
Explanation:
The five letters could be arrange in 5! ways.
One of them is 'BRING'.
? Required probability = 1/5!
= 1/(5 x 4 x 3 x 2 x 1)
= 1/120
- 10. If one rolls a fair sides die twice, what is the probability that the die will land on the same number on both the occasions?
Options- A. 1/12
- B. 1/6
- C. 1/24
- D. 1/8 Discuss
Correct Answer: 1/6
Explanation:
Let E = Event of getting same number on both the occasions
= (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)
? n(E) = 6
? Required probability = 6/62 = 1/6
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