BANK & INSURANCE (PROBABILITY) PART 2

Total Questions: 45

41. Daniel speaks truth in 2/5 cases and Sherin lies in 3/7 cases. What is the percentage of cases in which both Daniel and Sherin contradict each other in stating a fact?

Correct Answer: (d) 51.4%  
Solution:

Daniel and Sherin will contradict each other when one speaks truth and other speaks lies.

Probability of Daniel speak truth and Sherin lies
= 2/5 × 3/7 = 6/35

Probability of Sherin speak truth and Daniel lies
= 4/7 × 3/5 = 12/35

The two probabilities are mutually exclusive.

Hence, probabilities that Daniel and Sherin contradict each other:
= 6/35 + 12/35 = 18/35
= 18/35 × 100 = 51.4%

42. The probability that a bullet fired from a point will strike the target is 3/4. 5 such bullets are fired simultaneously towards the target from that very point. What is the probability that the target will be strike?

Correct Answer: (a) 255/256  
Solution:

Probability of a bullet not striking the target = 1/4

Probability that none of the 5 bullets will strike the target
= (1/4)⁵ = 1/256

Probability that the target will strike at least once:
= 1 − 1/256 = 255/256

43. Akshaya and Nikitha play a game where each is asked to select a number from 1 to 8. If the two numbers same, both of them win a prize. The probability that they will not win a prize in a single trial is:

Correct Answer: (d) 7/8  
Solution:

Total number of ways in which both of them can select a number each = 8×8 = 64
Total number of ways in which both of them can select a same number so that they both can win = 8 ways
[They both can select {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6), (7,7), (8,8)}]
Probability that they win the prize:
= Favorable Cases / Total Cases = 8/64 = 1/8
Probability that they do not win a prize = 1 − 1/8 = 7/8

44. There are three events X, Y and Z, one of which must and only can happen. If the odds are 7:4 against X, 5:3 against Y, the odds against Z must be:

Correct Answer: (a) 65/23  
Solution:

According to the question,
P(X’)/P(X) = 7/4
P(X’) = 7/11
P(X) = 4/11
P(Y’)/P(Y) = 5/3
P(Y’) = 5/8, P(Y) = 3/8

Now, out of X, Y and Z, one and only one can happen.

P(X) + P(Y) + P(Z) = 1
4/11 + 3/8 + P(Z) = 1
P(Z) = 1 − 4/11 − 3/8 = 88 − 32 − 33 / 88 = 23/88
P(Z’) = 1 − P(Z) = 1 − 23/88 = 65/88

So odd against Z
P(Z’)/P(Z) = 65/23

45. Six male and five female stands in queue for buy a cinema ticket. The probability that they stand in alternate positions is:

Correct Answer: (b) 1/462  
Solution:

Total number of possible arrangements for Six male and five female stand in queue = 11!

When they occupy alternate position the arrangement would be like:
MFMFMFMFMFM

Thus, total number of possible arrangements for males,
= 6 × 5 × 4 × 3 × 2

Total number of possible arrangements for females,
= 5 × 4 × 3 × 2

Required probability = 6 × 5 × 4 × 3 × 2 × 5 × 4 × 3 × 2 / (11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2)
= 1/462