BANK & INSURANCE (PROBABILITY) PART 3

Total Questions: 45

41. There are 40 bulbs in a bag. 40% of them are red. 50% of the remaining are blue and rest are yellow. If ___ bulbs are picked at random, then the probability that exactly one of the bulbs is blue is ___.

The values given in which of the following options will fill the blanks in the same order in which it is given to make the statement true:
I. 2, (28/65)  II. 3, (567/1235)  III. 1, (2/5)

Correct Answer: (d) Only I and II
Solution:

Number of red bulbs in the bag = 40 × 0.4 = 16
Number of blue bulbs = (40 − 16) × 0.2 = 12
Number of yellow bulbs = 40 − 12 − 16 = 12

For statement I:
Required probability = (¹²C₁ × ²⁸C₁)/(⁴⁰C₂)
= (28/65)
Therefore, I is true.

For statement II:
Required probability = (¹²C₁ × ²⁸C₂)/(⁴⁰C₃)
= (567/1235)
So, II is true.

For statement III:
Required probability = (12/40)
= (3/10)
Therefore, III is false.

Hence, option d.

42. In a biased coin, the probability of getting a tail is (1/8) more than the probability of getting a head. If this coin is tossed thrice, then what is the difference between the probability of getting 3 heads and the probability of getting 2 tails and a head?

Correct Answer: (a) (7/128)  
Solution:

Let the probability of getting a head = ‘x’
Then, probability of getting a tail = x + (1/8)
= (8x + 1)/8

So, we have x + {(8x + 1)/8} = (8x + 8x + 1)/8
= 1

Or, 16x + 1 = 8
So, x = (8 − 1)/16 = (7/16)

So, probability of getting a head and that of getting a tail is (7/16) and (9/16), respectively

On tossing the coin thrice, probability of getting 3 heads = (7/16)³ = (343/4096)

On tossing the coin thrice, probability of getting 2 tails and 1 head = (9/16)² × (7/16) = (567/4096)

So, difference in probability = {(567 − 343)/4096}
= (224/4096) = (7/128)

Hence, option a.

43. An advisory committee of 7 judges is to be selected from a group of 7 senior judges and 5 junior judges. In how many ways a committee having at least a senior and a junior judge can be selected?

Correct Answer: (b) 791  
Solution:

Case I: 6 senior and 1 junior judges are there in the committee
Number of ways = ⁷C₆ × ⁵C₁ = 7 × 5 = 35

Case II: 5 senior and 2 junior judges are there in the committee
Number of ways = ⁷C₅ × ⁵C₂ = 21 × 10 = 210

Case III: 4 senior and 3 junior judges are there in the committee
Number of ways = ⁷C₄ × ⁵C₃ = 35 × 10 = 350

Case IV: 3 senior and 4 junior judges are there in the committee
Number of ways = ⁷C₃ × ⁵C₄ = 35 × 5 = 175

Case V: 2 senior and 5 junior judges are there in the committee
Number of ways = ⁷C₂ × ⁵C₅ = 21 × 1 = 21

So the total number of ways = 35 + 210 + 350 + 175 + 21 = 791

Hence, option b.

44. A box contains ‘x + 5’ red, ‘x’ green and ‘x + 1’ blue balls. If the probability of getting a red ball is 1/9 more than that of getting a blue ball, then find the probability of getting a red, two green and a blue ball if four balls are randomly drawn from the box one after another and without replacement.

Correct Answer: (c) 15/119
Solution:

Total number of balls in the box = x + 5 + x + x + 1
= 3x + 6

According to question;
{ x + 5 − x − 1 }/(3x + 6) = 1/9

Or, 4/(x + 2) = 1/3
Or, 12 = x + 2
Or, x = 10

Number of red balls = x + 5 = 10 + 5 = 15
Number of green balls = x = 10
Number of blue balls = x + 1 = 11
Total number of balls = 15 + 10 + 11 = 36

Desired Probability = {¹⁵C₁ × ¹⁰C₂ × ¹¹C₁}/³⁶C₄
= (15 × 10 × 9 × 11 × 24)/(2 × 36 × 35 × 34 × 33)
= 15/119

Hence, option c.

45. 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

Hence, option d.