Discussion
Home ‣ Aptitude ‣ Probability Comments
- Question
- If three balls are picked at random, what is the probability that atleast one is red?
Options- A. 1/2
- B. 7/12
- C. 11/12
- D. 1/5
- Correct Answer
- 11/12
ExplanationTotal Number of outcomes = 10C3 = 120
Number of outcomes not containing red balls = 5C3 = 10
? Probability that at least one is red = 1 - 10/120 = 11/12 - 1. If two marbles are drawn at random, what is the probability that both are red?
Options- A. 3/7
- B. 1/2
- C. 2/11
- D. None of the these Discuss
- 2. If two caps are picked at random, what is the probability that atleast one is red?
Options- A. 1/3
- B. 16/21
- C. 19/33
- D. 7/19 Discuss
- 3. If three caps are picked at random, what is the probability that two are red and one is green?
Options- A. 9/22
- B. 6/19
- C. 1/6
- D. 3/22 Discuss
- 4. If four caps are picked at random, what is the probability that none is green?
Options- A. 7/99
- B. 5/99
- C. 7/12
- D. 5/12 Discuss
- 5. If two caps are picked at random, what is the probability that both are blue?
Options- A. 1/6
- B. 1/10
- C. 1/12
- D. None of these Discuss
- 6. If four balls are picked at random, what is the probability that two are green and two are blue?
Options- A. 1/18
- B. 1/70
- C. 3/5
- D. 1/2 Discuss
- 7. Two cards are drawn at random from a pack of 52 cards. What is the probability that the drawn cards are both aces?
Options- A. 1/221
- B. 2/13
- C. 2/26
- D. None of these Discuss
- 8. A card is drawn at random from a pack of 52 cards. A card is drawn at random. What is the probability that it is neither a heart nor a king?
Options- A. 4/13
- B. 9/13
- C. 2/13
- D. 4/13 Discuss
- 9. if from each of the three boxes containing 3 white and 1 black, 2 white and 2 black and 1 white and 3 black balls, one ball is drawn at random, then the probability that 2 white and 1 black ball will be drawn is?
Options- A. 13/32
- B. 1/4
- C. 1/32
- D. 3/16 Discuss
- 10. A box contains 3 white and 2 red balls. If we draw one ball and ball and without replacing the first ball. The probability of drawing red ball in the second drawn is?
Options- A. 8/25
- B. 2/5
- C. 3/5
- D. 21/25 Discuss
Probability problems
Search Results
Correct Answer: None of the these
Explanation:
Total number of possible outcomes
= n(S) = 12C2 = (12 x 11)/2 = 66
Total number of favorable events
= n(E) = 4C2
= (4 x 3) / (2 x 1) = 6
? Required probability = 6/66 = 1/11
Correct Answer: 19/33
Explanation:
Required probability = 1 - 8C2/ 12C2
= 1 - 28/66 = 38/66 = 19/33
Correct Answer: 3/22
Explanation:
Total number of caps = 12
? n(S) = 12C3 = 220
n(E1) = Out of 4 red caps, number of ways to pick 2 caps = 4C2 = 6
n(E2) = Out of 5 green caps
Number of ways to pick one cap = 5C1 = 5
P(E) = n(E1) x n(E2)/n(S) = (6 x 5)/220 = 3/22
Correct Answer: 7/99
Explanation:
Total number of caps = 12
Total number of result = n(S) = 12C4 = 495
Out of 5 caps, number of ways to not pick a green cap = n(E1) = 5C0 = 1
and out of 7 caps, number of ways to pic 4 caps = n(E) = 7C4 = 35
? Required probability = (1 x 35)/495 = 7/99
Correct Answer: None of these
Explanation:
Total number of caps = 2 + 4 + 5 + 1 = 12
Total number of outcomes = n(S) = 12 C2 = 66
Favorable number of outcomes = n(E) = 2C2 = 1
? Required probability = 1/66
Correct Answer: 1/70
Explanation:
Total number of outcomes = 10c4 = 210
Favourable number of outcomes = 3c2 x 2c2 = 3 x 1 = 3
? Required probability = 3/210 = 1/70
Correct Answer: 1/221
Explanation:
n(S) = Number of ways of drawing 2 cards out of 52
52C2 = (52 x 51) / (2 x 1) = 1326
n(E) = Number of ways of drawing 2 cards out of 4
= 4C2 = (4 x 3)/2 = 6
? P(E) = 6/1326 = 1/221
Correct Answer: 9/13
Explanation:
There are 13 hearts and 3 more kings
? P(heart or a king) = (13 + 3)/52 = 4/13
? P (neither a heart nor a king) = 1 - 4/13 = 9/13
Correct Answer: 13/32
Explanation:
You can have
W, W, B or W, B, W or B, W, W
Reqd. probability
= 3/4. 2/4. 3/4 + 3/4. 2/4. 1/4 + 1/4. 2/4. 1/4
= 9/32 + 3/32 + 1/32 = 13/32
Correct Answer: 2/5
Explanation:
Total balls in the box = 5
Second red ball can be drawn in two ways
Case I. First ball is white and second ball is red.
Its probability = 3/5.2/4 = 6/20 = 3/10
Case II: First ball is red and second ball is red
Its probability = 2/5.1/4 = 2/20 = 1/10
Hence, reqd. probability
= 3/10 + 1/10 = 4/10 = 2/5
Comments
There are no comments.More in Aptitude:
Programming
Copyright ©CuriousTab. All rights reserved.