BANK & INSURANCE (PROBABILITY) PART 2

Total Questions: 45

31. A bag contains 4 red balls and 4 yellow balls. If four balls are picked at random, then what is the probability that balls picked alternately are of different colour, if repetition not allowed?

Correct Answer: (b) 6/35  
Solution:

Required probability
= (⁴C₁/⁸C₁ × ⁴C₁/⁷C₁ × ³C₁/⁶C₁ × ³C₁/⁵C₁) + (⁴C₁/⁸C₁ × ⁴C₁/⁷C₁ × ³C₁/⁶C₁ × ³C₁/⁵C₁)
= ½ × 4/7 × ½ × 3/5 + ½ × 4/7 × ½ × 3/5
= 12/140 + 12/140
= 6/35

32. An urn consists of some green, red and black balls. Total number of balls in the urn is 15. If two balls are picked, Probability that the two balls being green is 2/21. Now, one ball is picked from the urn, find the probability that the ball is either red or black.

Correct Answer: (b) 2/15
Solution:

Let G, R and B be the number of green, red and black balls respectively.
G + R + B = 15

P (two balls are green) = 2/21
G × (G − 1)/(15 × 14) = 2/21

G² − G = 20
G² − G − 20 = 0
G² − 5G + 4G − 20 = 0
G(G − 5) + 4(G − 5) = 0
(G + 4)(G − 5) = 0
G = 5
R + B = 10
P (Either red or black) = 10/15 = 2/3

33. Two persons stole the gold chain. The probability of catching the first person is ¼. The probability of catching the second person is 1 / 6. What is the probability that only one of them is caught?

Correct Answer: (c) 1 / 3
Solution:

P(1st not caught) = 1 − 1/4 = 3/4
P(2nd not caught) = 1 − 1/6 = 5/6
P(Only one caught) = P(1st caught and 2nd not caught) + P(2nd caught and 1st not caught)
= (1/4 × 5/6) + (3/4 × 1/6) = 1/3

34. A bag contains (x + 3) pink, 4 blue and 6 white color balls. If two balls are taken random and the probability of getting both being blue color balls is 2/51, then find the total number of pink color balls?

Correct Answer: (c) 8 balls
Solution:

⁴C₂/(x + 13)C₂ = 2/51
[(4 × 3) / (1 × 2)] / [(x + 13)(x + 12) / (1 × 2)]
= 2/51

(51 × 6) = x² + 13x + 12x + 156
306 = x² + 25x + 156
x² + 25x − 150 = 0
(x − 5)(x + 30) = 0
x = 5, −30 (negative value will be eliminated)

The total number of pink color balls = 8 balls

35. There are three groups of fire extinguishers in a city i.e. P, Q and R. In group P there are (x-6) females and (x+21) males. In group Q, there are 3 females and 12 males whereas in group R, there are 5 females and 7 males. One person is selected at random from each group. Find the number of females in group P, if the probability of selecting all three males is 4/15.

Correct Answer: (a) 81  
Solution:

{(x + 21)/(x − 6 + x + 21)} × (12/15) × (7/12)
= 4/15

{(x + 21)/(2x + 15)} × (7/15) = 4/15
{(x + 21)/(2x + 15)} = 4/7

7x + 147 = 8x + 60
x = 87

Number of females in group P = 87 − 6 = 81

36. Two friends Harish and Kalyan appeared for an exam. Let A be the event that Harish is selected and B is the event that Kalyan is selected. The probability of A is 2/5 and that of B is 3/7. Find the probability that both of them are selected

Correct Answer: (d) 6/35  
Solution:

Given, A be the event that Harish is selected and
B is the event that Kalyan is selected.
P(A) = 2/5
P(B) = 3/7

Let C be the event that both are selected.
P(C) = P(A) × P(B) as A and B are independent events:
P(C) = 2/5 × 3/7 = 6/35

The probability that both of them are selected is 6/35

37. Preethi was born between August 26th and 30th (26th and 30th excluding). Her year of birth is also unknown. What is the probability of Preethi being born on a Monday?

Correct Answer: (d) 3/7  
Solution:

Since the year of birth is Unknown, the birthday being on Monday can have a zero probability.

Also since between 26th and 30th, there are three days
i.e. 27th, 28th and 29th

We have following possibilities on these three dates,
Monday, Tuesday, Wednesday
Tuesday, Wednesday, Thursday
Wednesday, Thursday, Friday
Thursday, Friday, Saturday
Friday, Saturday, Sunday
Saturday, Sunday, Monday
Sunday, Monday, Tuesday

Out of these 7 events, we have 3 chances of his birthday falling on Monday

Probability = favorable events / total events
= 3/7

Therefore, the probability of birthday falling on Monday can be 3/7.

38. A number is selected at random from first 40 natural numbers. What is the chance that it is a multiple of either 4 or 14?

Correct Answer: (b) 11/40  
Solution:

We know that,
Probability = Favorable Cases / Total Cases

The probability that the number is a multiple of 4 is
10/40

Since favorable cases here {4, 8, 12, 16, 20, 24, 28, 32, 36, 40}
= 10 cases

Total cases = 40 cases

Similarly the probability that the number is a multiple of 14 is 2/40.

Since favorable cases here {14, 28}
= 2 cases

Total cases = 40 cases

Multiple of 4 or 14 have common multiple from 1 to 40 is 28.

Hence, these events are mutually exclusive events.

Therefore chance that the selected number is a multiple of 4 or 14 is
= (10 + 2 − 1)/40 = 11/40

39. Srinaya forgot the last digit of an 11 digit land line number. If she randomly dials the final 2 digits after correctly dialing the first nine, then what is the chance of dialing the correct number?

Correct Answer: (b) 1/100  
Solution:

It is given that last two digits are randomly dialed.
Then each of the digits can be selected out of 10 digits in 10 ways.

Hence required probability
= (1/10)² = 1/100

40. A military man can strike a target once in 3 bullets. If he fires 3 bullets in succession, what is the probability that he will strike his target?

Correct Answer: (b) 19/27  
Solution:

The military man will strike the target even if he strike it once or twice or all three times in the three bullets that he takes.

So, the only case where the man will not strike the target is when he fails to strike the target even in one of the three bullets that he takes.

The probability that he will not strike the target in one bullets = 1 − Probability that he will strike target in exact one bullets
= 1 − 1/3 = 2/3

Therefore, the probability that he will not strike the target in all the three bullets
= 2/3 × 2/3 × 2/3
= 8/27

Hence, the probability that he will strike the target at least in one of the four bullets:
= (1 − 8/27) = 19/27