BANK & INSURANCE (PROBABILITY) PART 3

Total Questions: 45

21. Directions (21-23): Answer the questions based on the information given below.

A bag contains pink and green balls in the ratio 5:3, respectively. A box contains 3 white, 5 black and 4 green balls. Two balls are randomly drawn from the box and put into the bag. Now a ball is taken out of the bag, such that the probability that a green ball is drawn is 10/27.

Ques: Find the sum of the number of pink balls and green balls, in the bag initially.

Correct Answer: (b) 16  
Solution:

Let the bag contains 5x pink balls and 3x green balls.

Case I: Two green balls are drawn from the box.
The probability of drawing a green ball from the bag
= 4C₂/12C₂ × ((3x + 2)/(8x + 2)) = (1/11) × (3x + 2)/(8x + 2)

Case II: Only one green ball is drawn from the box.
The probability of drawing a green ball from the bag
= (4C₁ × 8C₁)/12C₂ × ((3x + 1)/(8x + 2))
= (16/33) × (3x + 1)/(8x + 2)

Case III: No green ball is drawn from the box:
The probability of drawing a green ball from the bag
= 8C₂/12C₂ × (3x/(8x + 2)) = (14/33) × (3x)/(8x + 2)

According to the question,
(1/11) × (3x + 2)/(8x + 2) + (16/33) × (3x + 1)/(8x + 2) + (14/33) × (3x)/(8x + 2) = 10/27

{1/11(8x + 2)} × {3x + 2 + (48x + 16)/3 + (42x/3)} = 10/27

= 10/27

x = 2

So, the total numbers of pink and green balls in the bag are 10 and 6, respectively.
Required sum = 10 + 6 = 16
Hence, option b.

22. If all the balls from the bag and box are mixed in a drum then find the probability of picking two green balls at a random from the drum.

Correct Answer: (d) 5/42  
Solution:

Let the bag contains 5x pink balls and 3x green balls.

Case I: Two green balls are drawn from the box.
The probability of drawing a green ball from the bag
= 4C₂/12C₂ × ((3x + 2)/(8x + 2)) = (1/11) × (3x + 2)/(8x + 2)

Case II: Only one green ball is drawn from the box.
The probability of drawing a green ball from the bag
= (4C₁ × 8C₁)/12C₂ × ((3x + 1)/(8x + 2))
= (16/33) × (3x + 1)/(8x + 2)

Case III: No green ball is drawn from the box:
The probability of drawing a green ball from the bag
= 8C₂/12C₂ × (3x/(8x + 2)) = (14/33) × (3x)/(8x + 2)

According to the question,
(1/11) × (3x + 2)/(8x + 2) + (16/33) × (3x + 1)/(8x + 2) + (14/33) × (3x)/(8x + 2) = 10/27

{1/11(8x + 2)} × {3x + 2 + (48x + 16)/3 + (42x/3)} = 10/27

x = 2

So, the total numbers of pink and green balls in the bag are 10 and 6, respectively.
Required probability = 10C₂/28C₂ = 5/42
Hence, option d.

23. If the number of pink balls in the bag had been twice the original number of pink balls and the number of green balls had been 2 less than the original number of green balls, then find the probability of picking 2 pink balls at a random from the bag.

Correct Answer: (a) 95/138  
Solution:

Let the bag contains 5x pink balls and 3x green balls.

Case I: Two green balls are drawn from the box.
The probability of drawing a green ball from the bag
= 4C₂/12C₂ × ((3x + 2)/(8x + 2)) = (1/11) × (3x + 2)/(8x + 2)

Case II: Only one green ball is drawn from the box.
The probability of drawing a green ball from the bag
= (4C₁ × 8C₁)/12C₂ × ((3x + 1)/(8x + 2))
= (16/33) × (3x + 1)/(8x + 2)

Case III: No green ball is drawn from the box:
The probability of drawing a green ball from the bag
= 8C₂/12C₂ × (3x/(8x + 2)) = (14/33) × (3x)/(8x + 2)

According to the question,
(1/11) × (3x + 2)/(8x + 2) + (16/33) × (3x + 1)/(8x + 2) + (14/33) × (3x)/(8x + 2) = 10/27

{1/11(8x + 2)} × {3x + 2 + (48x + 16)/3 + (42x/3)} = 10/27

Or, x = 2

So, the total numbers of pink and green balls in the bag are 10 and 6, respectively.

New number of pink balls = 10 × 2 = 20
New number of green balls = 6 − 2 = 4

Required probability = 20C₂/24C₂ = 95/138
Hence, option a.

24. A bag contains 4 green balls, 5 black balls, 3 red balls and 4 blue balls. Ram bets that he will pick two balls at random without replacement and both will be of the same colour. Shyam takes out all the blue balls from the bag and then bets that when he picks 2 balls, at random and without replacement, from the bag, both will be of different colours. Find the difference in probability between Ram and Shyam winning the bet.

Correct Answer: (d) (133/264)  
Solution:

Total number of balls in the bag initially = 4 + 5 + 3 + 4 = 16

Probability of picking 2 green balls at random
= 4C₂ ÷ 16C₂ = (6/120)

Probability of picking 2 black balls at random
= 5C₂ ÷ 16C₂ = (10/120)

Probability of picking 2 green balls at random
= 3C₂ ÷ 16C₂ = (3/120)

Probability of picking 2 blue balls at random
= 4C₂ ÷ 16C₂ = (6/120)

So, probability of Ram winning the bet
= (6 + 10 + 3 + 6) ÷ 120 = (25/120) = (5/24)

After picking out all the blue balls, total number of balls left in the bag = 4 + 5 + 3 = 12

Probability of Shyam picking two green balls
= 4C₂ ÷ 12C₂ = (6/66)

Probability of Shyam picking two black balls
= 5C₂ ÷ 12C₂ = (10/66)

Probability of Shyam picking two red balls
= 3C₂ ÷ 12C₂ = (3/66)

So, probability of Shyam picking two balls of the same colour
= (6 + 10 + 3) ÷ 66 = (19/66)

So, probability of Shyam picking two balls of different colours i.e. probability of Shyam winning the bet
= 1 − (19/66) = (47/66)

So, difference in probability = (47/66) − (5/24)
= (188/264) − (55/264)
= (133/264)

Hence, option d.

25. Directions (25-27): Answer the questions based on the information given below.

There are three bags viz. Bag A, Bag B, and Bag C containing 36, 32 and 48 balls respectively, each containing balls of three colours viz. Black, White and Grey. The ratio of number of Black balls to White balls in bag A and bag B is 3:2 and 2:3 respectively. The number of Grey balls in Bag A is equal to the number of White balls in bag C, and number of Black balls in bag C is three times to the number of Grey balls in bag C. The ratio of number of Grey balls in bag A to the number of Grey balls in bag B is 4:3 and ratio of number of Grey balls in bag B to the number of Grey balls in bag C is 3:2.

Ques : If ____ Pink balls are added to bag B and then a ball is randomly drawn from the bag, then the probability that a White ball is drawn is ____ .
The values given in which of the following options will fill the blanks in the same order in which is it given to make the above statement true:
A. 4, 1/3  B. 8, 3/11
C. 20, 3/13  D. 16, 1/4

Correct Answer: (d) Only A, C and D
Solution:

Let the number of Black balls and White balls in bag A be 3x and 2x respectively.
So the number of Grey balls in bag A
= 36 − (3x + 2x) = (36 − 5x)

Let the number of Black balls and White balls in bag B be 2y and 3y respectively.
So the number of Grey balls in bag B
= 32 − (2y + 3y) = (32 − 5y)

Let the number of Black balls and Grey balls in bag C be 3z and z respectively.
So the number of White balls in bag C
= (48 − 4z)

According to question,
36 − 5x = 48 − 4z
4z − 5x = 12 .......... (i)

Also, (36 − 5x)/(32 − 5y) = 4/3
108 − 15x = 128 − 20y
20y − 15x = 20
4y = 3x + 4 .......... (ii)

Also, (32 − 5y)/z = 3/2
64 − 10y = 3z .......... (iii)

Solving equations (i), (ii), and (iii), we get,
x = 4, y = 4, z = 8

So, bag A contains 12 Black, 8 White and 16 Grey balls, bag B contains 8 Black, 12 White and 12 Grey balls and bag C contains 24 Black, 16 White and 8 Grey balls.

For option A:
Probability that a White ball is drawn = 12/(32 + 4)
= 12/36 = 1/3
So, option A can be the answer.

For option B:
Probability that a White ball is drawn = 12/(32 + 8)
= 12/40 = 3/10
So, option B can't be the answer.

For option C:
Probability that a White ball is drawn = 12/(32 + 20)
= 12/52 = 3/13
So, option C can be the answer.

For option D:
Probability that a White ball is drawn = 12/(32 + 16)
= 12/48 = 1/4
So, option D can be the answer.

Hence, option d.

26. If two balls are randomly drawn from bag A then the probability that both the balls are Grey in colour is ____ ?

The values given in which of the following options will fill the blank to make the above statement true:

Correct Answer: (a) 4/21  
Solution:

Let the number of Black balls and White balls in bag A be 3x and 2x respectively,
So the number of Grey balls in bag A = 36 − (3x + 2x)
= (36 − 5x)

Let the number of Black balls and White balls in bag B be 2y and 3y respectively,
So the number of Grey balls in bag B = 32 − (2y + 3y)
= (32 − 5y)

Let the number of Black balls and Grey balls in bag C be 3z and z respectively,
So the number of White balls in bag C = (48 − 4z)

According to question,
36 − 5x = 48 − 4z
4z − 5x = 12 .......... (i)

Also, (36 − 5x)/(32 − 5y) = 4/3
108 − 15x = 128 − 20y
20y − 15x = 20
4y = 3x + 4 .......... (ii)

Also, (32 − 5y)/z = 3/2
64 − 10y = 3z .......... (iii)

Solving equations (i), (ii), and (iii), we get,
x = 4, y = 4, z = 8

So, bag A contains 12 Black, 8 White and 16 Grey balls, bag B contains 8 Black, 12 White and 12 Grey balls and bag C contains 24 Black, 16 White and 8 Grey balls.

Therefore, required probability = 16C₂/36C₂ = 4/21

Hence, option a.

27. A ball is transferred from bag A to bag C and then a ball is drawn from bag C. Find the probability that a White ball is drawn from bag C if it is known that the ball which is transferred from bag A to bag C is either Black or White.

Correct Answer: (d) 82/441  
Solution:

Let the number of Black balls and White balls in bag A be 3x and 2x respectively,
So the number of Grey balls in bag A = 36 − (3x + 2x)
= (36 − 5x)

Let the number of Black balls and White balls in bag B be 2y and 3y respectively,
So the number of Grey balls in bag B = 32 − (2y + 3y)
= (32 − 5y)

Let the number of Black balls and Grey balls in bag C be 3z and z respectively,
So the number of White balls in bag C = (48 − 4z)

According to question,
36 − 5x = 48 − 4z
4z − 5x = 12 .......... (i)

Also, (36 − 5x)/(32 − 5y) = 4/3
108 − 15x = 128 − 20y
20y − 15x = 20
4y = 3x + 4 .......... (ii)

Also, (32 − 5y)/z = 3/2
64 − 10y = 3z .......... (iii)

Solving equations (i), (ii), and (iii), we get,
x = 4, y = 4, z = 8

So, bag A contains 12 Black, 8 White and 16 Grey balls, bag B contains 8 Black, 12 White and 12 Grey balls and bag C contains 24 Black, 16 White and 8 Grey balls.

Case I: If a Black ball is transferred from bag A
So the probability that a White ball is drawn
= (12/36) × (16/49) = 16/147

Case II: If a White ball is transferred from bag A
So the probability that a White ball is drawn
= (8/36) × (17/49) = 34/441

Probability that ball drawn is White in colour
= (16/147) + (34/441) = (48 + 34)/441
= 82/441

Hence, option d.

28. A bag contains ‘x’ blue balls, (x - 2) red balls and (x - 1) green balls. If two balls are picked at random without replacement from the bag, then the probability that both balls are blue balls is (5/33).

Which among the given statement(s) is/are true based on the statement given above?

I. The probability of picking 3 balls at random without replacement and getting all 3 as red balls is (1/220)
II. Total number of balls in the bag is at least 10.
III. If all blue balls are removed from the bag, and then 2 balls are picked at random without replacement, then the probability of getting 2 balls of different colour is (1/2)

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

Total number of balls in the bag
= x + (x − 2) + (x − 1) = (3x − 3)

Probability of picking 2 blue balls at random
= xC₂ ÷ (3x − 3)C₂
= {(x × (x − 1)) ÷ 2} ÷ {[(3x − 3)(3x − 4)] ÷ 2}
= (x × x) ÷ (9x² − 9x − 12x + 12) = (5/33)

Or, 33x² − 33x = 45x² − 45x − 60x + 60
Or, 12x² − 72x + 60 = 0
Or, 12x² − 12x − 60x + 60 = 0
Or, 12x(x − 1) − 60(x − 1) = 0
Or, (12x − 60)(x − 1) = 0

So, x = 5 or 1
But since (x − 2) will be negative for x = 1, so x = 5

So, number of red balls, blue balls and green balls in the bag are 3, 5 and 4, respectively.
And, total number of balls in the bag = 3 + 5 + 4
= 12

For I:
Probability of picking 3 red balls at random = ³C₃ ÷ ¹²C₃ = (1/220)
So, statement I is true.

For II:
Total number of balls in the bag = 12 ≥ 10
So, statement II is true.

For III:
After removing all blue balls, total number of balls in the bag = 3 + 4 = 7
Probability of picking two red balls at random from this bag = ³C₂ ÷ ⁷C₂ = (1/7)
Probability of picking two green balls at random from this bag = ⁴C₂ ÷ ⁷C₂ = (2/7)

So, required probability = 1 − {(1/7) + (2/7)}
= 1 − (3/7) = (4/7) ≠ (1/2)

So, statement III is false.
Hence, option b.

29. Three consecutive letters are selected at random from the word ‘ADVENTURE’. What is the probability that at least one is a vowel?

Correct Answer: (c) 0.8
Solution:

Possible pairs of three consecutive letters = (ADV, DVE, VEN, ENT, NTU, TUR, URE)

We can see that all of these combinations, contain 1 or more vowel, hence the required probability is 7/9
= .8

Hence, option c.

30. A bag contains green and grey balls in the ratio 5:6 respectively. A box contains 4 red, 6 blue and 5 green balls. Two balls are randomly drawn from the box and put in the bag. Now a ball is randomly drawn from the bag and the probability that the ball drawn is green is 4/9. Find the total number of grey balls in the bag.

Correct Answer: (b) 12  
Solution:

Let the bag contains 5x green and 6x grey balls.

Case I: Two green balls are drawn from the box
The probability of drawing a green ball from the bag
= ⁵C₂/¹⁵C₂ × (5x + 2)/(11x + 2) = 2/21 × (5x + 2)/(11x + 2)

Case II: Only one green ball is drawn from the box
The probability of drawing a green ball from the bag
= ⁵C₁ × ¹⁰C₁/¹⁵C₂ × (5x + 1)/(11x + 2)
= 10/21 × (5x + 1)/(11x + 2)

Case III: No green ball is drawn from the box
The probability of drawing a green ball from the bag
= ¹⁰C₂/¹⁵C₂ × 5x/(11x + 2) = 9/21 × (5x)/(11x + 2)

So according to question:
2/21 × (5x + 2)/(11x + 2) + 10/21 × (5x + 1)/(11x + 2) + 9/21 × (5x)/(11x + 2) = 4/9

1/(21(11x + 2)) × (10x + 4 + 50x + 10 + 45x) = 4/9

1/(7(11x + 2)) × (105x + 14) = 4/3

315x + 42 = 308x + 56
7x = 14
x = 2

So the total number of grey balls in the bag = 2 × 6 = 12

Hence, option b.