BANK & INSURANCE (PROBABILITY) PART 1

Total Questions: 45

21. Bharti speaks truth in 70% cases and Amit speaks truth in 40% cases. The probability that they will say the same thing while describing an event is?

Correct Answer: (e) None of these
Solution:P (Bharti speaking truth) = 0.7, thus, P (Bharti lying) = 0.3
P (Amit speaking truth) = 0.4, thus, P (Amit lying) = 0.6
Probability of saying the same thing is when both speak truth or lie together. Therefore, required probability = P (Bharti speaking truth) × P (Amit speaking truth) + P (Bharti lying) × P (Amit lying)
= i.e. 0.7 × 0.4 + 0.6 × 0.3 = 0.46

22. Twelve applicants apply for a job of which 5 are women and 7 are men. It is desired to select 3 persons for the job. What is the probability that at least one of the selected person is a woman?

Correct Answer: (c) 37/44
Solution:Total Number of ways in which 3 persons can be selected out of 12 persons is ¹²C₃
n(E) = (⁵C₁ × ⁷C₂) + (⁵C₂ × ⁷C₁) + ⁵C₃ = 105 + 70 + 10 = 185
n(S) = ¹²C₃ = 220
P(E) = 185/220 = 37/44

23. Seven applicants are applying for the Campaign Manager team of Shaastra, but only three spots are vacant. There are 5 girls and 2 boys applying for the post. What is the probability that the three vacant spots are filled by at least one boy?

Correct Answer: (d) 5/7  
Solution:Total Number of ways in which 3 spots can be filled by 7 applicants is ⁷C₃
n(E) = (²C₁ × ⁵C₂) + (⁵C₁ × ²C₂) = 25
n(S) = ⁷C₃ = 35
P(E) = 25/35 = 5/7

24. A bag contains 6 Black balls and 4 White balls. A second bag contains 5 Black balls and 3 White balls. One bag is selected at random. From the selected bag one ball is drawn. What is the probability that the drawn ball is white?

Correct Answer: (a) 31/80  
Solution:Probability of selecting a bag = 1/2
Probability of selecting white ball from the first bag (p(E₁)) = 4/10 = 2/5
Probability of selecting white ball from the second bag (p(E₂)) = 3/8
Total probability = p(E₁) + p(E₂) = 2/5 + 3/8 = 31/40
Required probability p(E) = 1/2 × 31/40 = 31/80

25. From a well shuffled deck of playing cards, 3 cards are drawn at random. What is the probability that they are Ace, King or Queen of same color?

Correct Answer: (b) 2/1105  
Solution:

1st card can be drawn in 12/52 ways (4 Ace, 4 King & 4 Queen)
2nd card can be drawn in 5/51 ways
Suppose Diamond Ace is picked as a first card, then we have choices for Heart Ace, 2 King & 2 Queens, so total 5 choices out of 51 cards
3rd card can be drawn in 4/50 ways
Hence, probability that all three cards are Ace, King & Queen of same colour = 12/52 × 5/51 × 4/50 = 2/1105

26. After a typist had written thirty letters and had addressed the thirty corresponding envelopes, a careless mailing clerk inserted the letters in the envelopes at random, one letter per envelope. What is the probability that exactly twenty nine letters are inserted in proper envelope?

Correct Answer: (e) None of these
Solution:No such case would exist such that after putting twenty nine letters in proper envelope, 30th letter is placed in a wrong envelope. Thus, the probability of such an event is zero.

27. Find out the probability of forming 1359 or 1273 with the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 when only four digit numbers are formed and when repetitions are not allowed?

Correct Answer: (b) 1/1512  
Solution:Positive outcomes = 2 (1359 or 1273)
Total outcomes = 9 × 8 × 7 × 6 = 3024
Required probability = 2/3024 = 1/1512

28. There are 3 bags and 5 keys. In how many ways the keys can be distributed in the bags such that all the keys are identical and 2 bags out of the 3 are identical and looks the same?

Correct Answer: (e) None of these
Solution:

We have to take cases to solve this

  1. (5,0,0): In this case all 5 keys can be placed in either the unique bag or the identical one. So 2 cases possible.
  2. (4,1,0): This can have 3 cases as the unique bag can have either 4,1 or 0 keys.
  3. (3,2,0): This can also have 3 cases.
  4. (2,2,1): This can have 2 cases as the unique bag can have 2 or 1 keys.
  5. (3,1,1): This can have 2 cases as well.

So, total number of ways = 2 + 3 + 3 + 2 + 2 = 12 ways

29. In how many ways can 4 boys and 3 girls be arranged in a line such that a group of 3 of the boys and another group of 2 of the girls are always together and the groups of the 3 boys and the 2 girls are never adjacent to each other?

Correct Answer: (d) 144  
Solution:3 boys can be arranged amongst themselves in 3! = 6 ways and the 2 girls can be arranged amongst themselves in 2! = 2 ways. Remaining one boy and one girl can be arranged in 2 ways and the group of boys and girls can be arranged in 2 ways. Number of ways per arrangement = 6 × 2 × 2 × 2 = 48 ways
Let the group of boys be B and the group of girls be G.
Case I: B/G _ G/B Number of ways = 48
Case II: B/G _ G/B _ Number of ways = 48
Case III: _ B/G _ G/B Number of ways = 48
Total number of ways = 48 × 3 = 144 ways

30. How many words can be formed from the letters of the word “CONSUMER” with or without meaning, so that all the vowels are never together?

Correct Answer: (b) 36000  
Solution:The word ‘CONSUMER’ contains 8 different letters
Taking the vowels OUE together & treating them as one letter, then all the letters can be arranged in 6! ways = 720 ways
The vowels OUE can be arranged among themselves in 3! ways = 6 ways
So, number of words formed, each having vowels together = 720 × 6 = 4320
Total number of words formed by using all the 8 letters of the given word = 8! = 40320
So, number of words each having vowels never together = 40320 − 4320 = 36000