Probability problems
- 1. The odds against the occurrence of an event are 5 : 4. The probability of its occurrence is?
Options- A. 4/5
- B. 4/9
- C. 1/5
- D. 1/4 Discuss
Correct Answer: 4/9
Explanation:
Number of cases favourable of E = 4
Total Number of cases = (5 + 4 ) = 9
? P(E) = 4/9
- 2. A bag contains 8 red and 5 white balls. 2 balls are drawn at random. What is the probability that both are white?
Options- A. 5/16
- B. 2/13
- C. 3/26
- D. 5/39 Discuss
Correct Answer: 5/39
Explanation:
n(S) = Number of ways of drawing 2 balls out of 13 = 13C2 = 13 x 12 / 2= 78
n(E) = No. of ways of drawing 2 balls out of 5 = 5C2 = 5 x 4 / 2 = 10
? P(E) = n(E)/n(S) = 10/78 = 5/39
- 3. A bag contains 6 black balls and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white?
Options- A. 4/7
- B. 3/4
- C. 4/5
- D. 1/8 Discuss
Correct Answer: 4/7
Explanation:
Total no. of balls = (6 + 8) = 14
No. of white balls = 8
? P(drawing a white ball) = 8/14 = 4/7
- 4. Three unbiased coins are tossed. What is the probability of getting at most 2 heads?
Options- A. 1/4
- B. 3/8
- C. 7/8
- D. 1/2 Discuss
Correct Answer: 7/8
Explanation:
? n(S) = (2)3 = 8
E = Event of getting 0, or 1 or 2 heads
= {TTT, TTH, THT, HTT, HHT, HTH, THH}
? n(E) = 7
? P(E) = n(E)/n(S) = 7/8
- 5. Three unbiased coins are tossed, what is the probability of getting exactly two heads?
Options- A. 1/3
- B. 3/4
- C. 2/3
- D. 3/8 Discuss
Correct Answer: 3/8
Explanation:
S= { HHH, HHT, HTH, THH, TTH, THT, HTT, TTT} and
E = Event of getting exactly two heads = {HHT, HTH, THH}
? P(E) = n(E)/n(S) = 3/8
- 6. In a simultaneous throw of two coins, the probability of getting at least one head is?
Options- A. 1/2
- B. 2/3
- C. 3/4
- D. 1/3 Discuss
Correct Answer: 3/4
Explanation:
S = {HH, HT, TT, TH}
and E = {HH, HT, TH }
? P(E) = n(E)/n(S) = 3/4
- 7. In a throw of a coin, the probability of getting a head is?
Options- A. 1/2
- B. 1/4
- C. 1
- D. None of these Discuss
Correct Answer: 1/2
Explanation:
Here S = {H, T}
and E = {H}
? P(E) = n(E)/n(S) = 1/2
- 8. The probability that a teacher will give one surprise test during any class meeting in a week is 1/5. If a student is absent twice. What is the probability that he will miss at least one test?
Options- A. 4/15
- B. 1/15
- C. 91/25
- D. 16/125 Discuss
Correct Answer: 1/15
Explanation:
The probability of absenting of the students in the class = 2/6 = 1/3
? The probability of missing his test = 1/5 x 1/3 = 1/15
- 9. A coin is successively tossed two times. Find the probability of getting? (1) exactly one head (2)at least one head
Options- A. 1/2 , 3/4
- B. 2/3, 1/4
- C. 1/4, 4/5
- D. 1/2, 2/3 Discuss
Correct Answer: 1/2 , 3/4
Explanation:
In tossing a coin 2 times the sample space is 4 i,e (H, H), (H, T), (T, H), (T, T)
(1) If A1 denotes exactly one head
then A1 = {(H, T) (T, H) }
So, P(A1 ) = 2/4 = 1/2
(2) If A denotes at least one head
then A = {(H, T) (T, H) (H, H)}
? A = {(H, T) (T, H) (H, H )}
? P(A) = 3/4
- 10. In a simultaneous throw of two dice find the probability of getting a total of 8 .
Options- A. 2/9
- B. 5/36
- C. 1/6
- D. Data Inadequate Discuss
Correct Answer: 5/36
Explanation:
In a simultaneous throw of two dice
Simple Space = 6 x 6 = 36
Favorable cases are = (2, 6) (3, 5) (4, 4 ) (5, 3) (6, 2)
So, The required probability = 5/36
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