Discussion
Home ‣ Aptitude ‣ Probability Comments
- Question
- A six - faced dice is so biased that is twice as likely to show an even number as an odd number when throw. It is thrown twice. The probability that the sum of two number thrown is even is?
Options- A. 1/12
- B. 1/6
- C. 1/3
- D. 5/9
- Correct Answer
- 5/9
Explanation? Probability for odd = p
? Probability for even = 2p
? p + 2p = 1
? 3p = 1
? p = 1/3
? Probability for odd = 1/3, Probability for even = 2/3.
Sum of two nos. is even means either both are odd or both are even
? Reqd. probability = 1/3 x 1/3 + 2/3 x 2/3 = 1/9 + 4/9 = 5/9
[ ? die is thrown twice ] - 1. A bag contains 5 blue and 4 black balls. Three balls are drawn at random. What is the probability that 2 are blue and 1 is black?
Options- A. 1/3
- B. 2/5
- C. 1/6
- D. None of these Discuss
- 2. Two number are selected randomly from the set S = {1, 2, 3, 4, 5, 6} without replacement one by one. The probability that minimum of the number is less than 4 is?
Options- A. 1/15
- B. 14/15
- C. 1/5
- D. 4/5 Discuss
- 3. The probability of occurrence of a multiple of 2 on a dice and multiple of 3 on the other dice . If both are thrown together is?
Options- A. 7/26
- B. 1/3
- C. 71/36
- D. 1/4 Discuss
- 4. Seven white balls and three black ball are randomly placed in a row. The probability that no two black balls are placed adjacently equals?
Options- A. 1/2
- B. 7/15
- C. 2/15
- D. 1/3 Discuss
- 5. If the probability of machine failing during a day is 0.95. The probability of its working for four consecutive days without failing is?
Options- A. 0.00000625
- B. 0.0625
- C. 0.16548375
- D. 0.000625 Discuss
- 6. The probability of occurrence of an event A is 5/9 . The probability of non-occurrence of the event B is 5/11. The probability that at least one of them will occur?
Options- A. 6/11
- B. 5/9
- C. 4/9
- D. 0.8 Discuss
- 7. When two dice are thrown, the probability that the difference of the number on the dice is 2 or 3 is?
Options- A. 7/18
- B. 3/11
- C. 5/18
- D. 1/2 Discuss
- 8. From a set of 17 cards numbered 1, 2, 3 ....... 17 one is drawn at random. The probability that the number is divisible by 3 or 7 is?
Options- A. 2/17
- B. 1/7
- C. 7/17
- D. 10/ 17 Discuss
- 9. A card is drawn at random from a pack of 100 cards numbered 1 to 100. The probability drawing a number which is a square is?
Options- A. 1/5
- B. 2/5
- C. 1/10
- D. None of these Discuss
- 10. In suffling a pack of card 3 are accidentally dropped then the chance that missing card should be of different suit is?
Options- A. 169/425
- B. 261/425
- C. 104/425
- D. None of these Discuss
Probability problems
Search Results
Correct Answer: 1/6
Explanation:
Let S be the sample space and E be the event of drawing 3 balls out of which 2 are blue and 1 is back .
Then, n (S)= Number of ways of drawing 3 balls out of 9=9c3
= (9 x 8 x 7) / (3 x 2 x 1)
= 84 and
n(E) = Number of ways of drawing 2 balls out of 5 and 1 ball out of 4 .
= 5C2 4C11
= (5 x 4) / (2 x 1) + 4 = 14
? P(E) = n(E)/n(S) = 14/84 = 1/6
Correct Answer: 4/5
Explanation:
Total no . of cases = 6P2 = 6 x 5 = 30
Non-favourable cases are (4, 5) , (5, 4), (4, 6), (6, 4), (5, 6), (6, 5)
? Probability that event will not happen = 6 /(6 x 5) = 1/5
? Reqd. probability = 1 - 1/5 = 4/5
Correct Answer: 1/3
Explanation:
Probability of multiple of 2 = 3/6 = 1/2
Probability of multiple of 3 = 2/6 = 1/3
Since there are two dice.
? The required probability = 2 x 1/2 x 1/3 = 1/3
Correct Answer: 7/15
Explanation:
n(S) = 10C7 = 10C3
n(E) = 8C3
? P(E) = n(E)/n(S) = 7/15
Correct Answer: 0.0625
Explanation:
? Probability of machine failing during a day p = 0.95
? q = Probability of its working during a day = 1 - p = 1 - 0.95 = 0.05
Required probability = q4 = (0.05)4
= 0.00000625
Correct Answer: 0.8
Explanation:
? P(A) = 5/9
? P(A) = 1- 5/9 = 4/9
? P(B) = 5/11
? P(B) = 1 - 5/11 = 6/11
Probability that none of them will occur = P(A ? B) = 4/9 x 5/ 11 = 20/99
Hence, Reqd. probability = 1-20/99 = 79//99
= 0.798 which is near 0.8.
Correct Answer: 7/18
Explanation:
The favourable cases are (1, 3), (2, 4) , (3, 5), (4, 6) and (1, 4), (2, 5), (3, 6) and their reversed cases like (3, 1), (4, 2), (5, 3)...
Total number of favourable cases = 2 x 7
? Required Probability P(E) = 14/36 = 7/18
Correct Answer: 7/17
Explanation:
Total number of cases is 17.
? Number divisible by 3 are 3, are 3, 6, 9, 12, 15 (These are 5 in number )
Number divisible by 7 are 7, 14. (These are 2 in number )
There are two favourable number of cases
Total no. of favourable number = 5 + 2
Required probability = 7/17.
Correct Answer: 1/10
Explanation:
Total ways = 100
Squares of following no's lie between 1 and 100,
12, 22, 32 , 42, 52 , 62, 72, 82 , 92 , 102
(Which are 10 in numbers.)
So. Required probability = 10/100 = 1/10
Correct Answer: 169/425
Explanation:
Total ways 52C3 = 22100
There are 4 suit in a pack of cards so three suit can be selected in 4C3 ways.
One card each from different unit can selected as = 13C1 X 13C1 X 13C1 ways
So, favourable ways = 4C3 X 13C1 X 13C1 X 13C1 = 8788
? Required probability = 8788/22100 = 169/425
Comments
There are no comments.More in Aptitude:
Programming
Copyright ©CuriousTab. All rights reserved.