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- Question
A coin is tossed 5 times. What is the probability that head appears an odd number of times?
Options- A. 1/2
- B. 1/3
- C. 2/3
- D. 1
- Correct Answer
- 1/2
ExplanationThe possible outcomes are as follows :
5H, 5T, (H, 4T), (T, 4H), (2H, 3T) (3H, 2T), i.e. 6 outcomes in all.
Therefore the probability that head appears an odd number of times = 3/6 =1/2 (In only three outcomes out of the six outcomes, head appears an odd number of times).
- 1. When two dice are thrown simultaneously, what is the probability that the sum of the two numbers that turn up is less than 12?
Options- A. 35/36
- B. 17/36
- C. 15/36
- D. 1/36 Discuss
- 2. A bag contains 3 red balls, 5 yellow balls and 7 pink balls. If one ball is drawn at random from the bag, what is the probability that it is either pink or red?
Options- A. 2/3
- B. 1/8
- C. 3/8
- D. 3/4 Discuss
- 3. Two cards are drawn from a pack of well shuffled cards. Find the probability that one is a club and other in King
Options- A. 1/52
- B. 1/26
- C. 1/13
- D. 1/2 Discuss
- 4. Two brother X and Y appeared for an exam. The probability of selection of X is 1/7 and that of B is 2/9. Find the probability that both of them are selected.
Options- A. 1/63
- B. 1/14
- C. 2/63
- D. 1/9 Discuss
- 5. An urn contains 4 white 6 black and 8 red balls . If 3 balls are drawn one by one without replacement, find the probability of getting all white balls.
Options- A. 5/204
- B. 1/204
- C. 13/204
- D. None of these Discuss
- 6. A five digit number is formed with the digits 0,1,2,3 and 4 without repetition. Find the chance that the number is divisible by 5.
Options- A. 3/5
- B. 1/5
- C. 2/5
- D. 4/5 Discuss
- 7. What is the probability of an impossible event?
Options- A. 0
- B. -1
- C. 0.1
- D. 1 Discuss
- 8. What is the probability of getting at least one six in a single throw of three unbiased dice?
Options- A. 1/36
- B. 91/256
- C. 13/256
- D. 43/256 Discuss
- 9. Ajay and his wife Reshmi appear in an interview for two vaccancies in the same post. The Probability of Ajay's selection is 1/7 and that of his wife Reshmi's selection is 1/5. What is the probability that only one of them will be selected?
Options- A. 5/7
- B. 1/5
- C. 2/7
- D. 2/35 Discuss
- 10. If a box contains 10 bulbs,of which just three are defective. If a random sample of five bulbs is drawn, find the probability that the sample contains no defective bulbs.
Options- A. 5/12
- B. 7/12
- C. 3/14
- D. 1/12 Discuss
Probability problems
Search Results
Correct Answer: 35/36
Correct Answer: 2/3
Explanation:
Given number of balls = 3 + 5 + 7 = 15
One ball is drawn randomly = 15C1
probability that it is either pink or red =
Correct Answer: 1/26
Explanation:
Let X be the event that cards are in a club which is not king and other is the king of club.
Let Y be the event that one is any club card and other is a non-club king.
Hence, required probability:
=P(A)+P(B)
=
= = =
Correct Answer: 2/63
Explanation:
Let A be the event that X is selected and B is the event that Y is selected.
P(A) = 1/7, P(B) = 2/9.
Let C be the event that both are selected.
P(C) = P(A) × P(B) as A and B are independent events:
= (1/7) × (2/9) = 2/63
Correct Answer: 1/204
Explanation:
Let A, B, C be the events of getting a white ball in first, second and third draw respectively, then
Required probability =
=
Now, P(A) = Probability of drawing a white ball in first draw = 4/18 = 2/9
When a white ball is drawn in the first draw there are 17 balls left in the urn, out of which 3 are white
Since the ball drawn is not replaced, therefore after drawing a white ball in the second draw there are 16 balls left in the urn, out of which 2 are white.
Hence the required probability =
Correct Answer: 1/5
Explanation:
Correct Answer: 0
Explanation:
The probability of an impossible event is 0.
The event is known ahead of time to be not possible, therefore by definition in mathematics, the probability is defined to be 0 which means it can never happen.
The probability of a certain event is 1.
Correct Answer: 91/256
Explanation:
Find the number of cases in which none of the digits show a '6'.
i.e. all three dice show a number other than '6', 5×5×5=125 cases.
Total possible outcomes when three dice are thrown = 216.
The number of outcomes in which at least one die shows a '6' = Total possible outcomes when three dice are thrown - Number of outcomes in which none of them show '6'.
=216?125=91
The required probability = 91/256
Correct Answer: 2/7
Explanation:
P( only one of them will be selected) = p[(E and not F) or (F and not E)]
=
=
=
Correct Answer: 1/12
Explanation:
Total number of elementary events =
Number of ways of selecting no defective bulbs i.e., 5 non-defective bulbs out of 7 is .
So,required probability = / = 1/12.
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