BANK & INSURANCE (PROBABILITY) PART 3

Total Questions: 45

1. Directions (1-2): Answer the questions based on the information given below.

There are two bags ‘A’ and ‘B’ and both contain certain number of red, green and blue balls. Number of red balls in bag ‘B’ is three less than that in bag ‘A’ which contains two more blue balls than the bag ‘B’. Number of green balls in bag ‘A’ is respectively one more and four more than number of blue and red balls in the same bag. When two balls are drawn one after another and without replacement from bag ‘B’ then probability of getting a green and a blue ball is (1/3). If balls in bag ‘A’ and bag ‘B’ are mixed together and a green ball is drawn then probability of getting a green ball is (3/8).

Ques: If number of balls in bag ‘A’ and ‘B’ are mixed together and two balls are drawn from it one after another randomly and without replacement, then find the probability of getting a green and a blue ball.

Correct Answer: (a) 4/13  
Solution:

Let number of red balls in bag ‘A’ be ‘x’
So, number of green balls in bag ‘A’ = x + 4
Number of blue balls in bag ‘A’ = x + 4 − 1 = x + 3
Number of red balls in bag ‘B’ = x − 3
Number of blue balls in bag ‘B’ = x + 3 − 2 = x + 1
Let number of green balls in bag ‘B’ be ‘y’

Total number of balls in bag ‘A’ = x + x + 3 + x + 4
= 3x + 7

Total number of balls in bag ‘B’ = x − 3 + y + x + 1
= 2x + y − 2

According to question;
(x + 4 + y)/(5x + y + 5) = 3/8
8x + 8y + 32 = 15x + 3y + 15
7x − 5y = 17 ...(1)

And, {2y(x + 1)}/[(2x + y − 2)(2x + y − 3)] = 1/3

Putting y = (7x − 17)/5, we get
79x² − 703x + 1374 = 0
79x² − 474x − 229x + 1374 = 0
79x(x − 6) − 229(x − 6) = 0
(x − 6)(79x − 229) = 0
x = 6 or x = 229/79 (not possible)

So, x = 6
And, y = (7×6 − 17)/5 = 5

Number of red balls in bag ‘A’ = x = 6
Number of green balls in bag ‘A’ = x + 4 = 6 + 4 = 10
Number of blue balls in bag ‘A’ = x + 3 = 6 + 3 = 9

Number of red balls in bag ‘B’ = x − 3 = 6 − 3 = 3
Number of blue balls in bag ‘B’ = x + 1 = 6 + 1 = 7
Number of green balls in bag ‘B’ = y = 5

Total number of balls in bag ‘A’ = 3x + 7 = 3×6 + 7 = 25
Total number of balls in bag ‘B’ = 2x + y − 2 = 2×6 + 5 − 2 = 15

Total number of red balls = 6 + 3 = 9
Total number of green balls = 10 + 5 = 15
Total number of blue balls = 9 + 7 = 16

Total number of balls = 9 + 15 + 16 = 40

Desired probability = (¹⁵C₁ × ¹⁶C₁) / ⁴⁰C₂ = 4/13
Hence, option a.

2. Total number of balls in bag ‘A’ is:

Correct Answer: (c) 25
Solution:

Let number of red balls in bag ‘A’ be ‘x’
So, number of green balls in bag ‘A’ = x + 4
Number of blue balls in bag ‘A’ = x + 4 − 1 = x + 3
Number of red balls in bag ‘B’ = x − 3
Number of blue balls in bag ‘B’ = x + 3 − 2 = x + 1
Let number of green balls in bag ‘B’ be ‘y’

Total number of balls in bag ‘A’ = x + x + 3 + x + 4
= 3x + 7

Total number of balls in bag ‘B’ = x − 3 + y + x + 1
= 2x + y − 2

According to question;
(x + 4 + y)/(5x + y + 5) = 3/8
8x + 8y + 32 = 15x + 3y + 15
7x − 5y = 17 ...(1)

And, {2y(x + 1)}/[(2x + y − 2)(2x + y − 3)] = 1/3

Putting y = (7x − 17)/5, we get
79x² − 703x + 1374 = 0
79x² − 474x − 229x + 1374 = 0
79x(x − 6) − 229(x − 6) = 0
(x − 6)(79x − 229) = 0
x = 6 or x = 229/79 (not possible)

So, x = 6
And, y = (7×6 − 17)/5 = 5

Number of red balls in bag ‘A’ = x = 6
Number of green balls in bag ‘A’ = x + 4 = 6 + 4 = 10
Number of blue balls in bag ‘A’ = x + 3 = 6 + 3 = 9

Number of red balls in bag ‘B’ = x − 3 = 6 − 3 = 3
Number of blue balls in bag ‘B’ = x + 1 = 6 + 1 = 7
Number of green balls in bag ‘B’ = y = 5

Total number of balls in bag ‘B’ = 2x + y − 2 = 2×6 + 5 − 2 = 15

Total number of balls in bag ‘A’ = 3x + 7 = 3×6 + 7 = 25

Hence, option c.

3. There are two bags ‘A’ and ‘B’ and both of them contain certain number of yellow, white and black caps. Number of yellow, white and black caps in bag ‘A’ is ‘x - 4’, 9 and ‘x - 1’ respectively. If two caps are drawn from bag ‘A’ one after another and without replacement then probability of getting a yellow and a black cap is (2/7). Find the number of white caps in bag ‘B’ if total number of yellow caps in bags ‘A’ and ‘B’ together is 20 and number of black caps in bag ‘A’ is 66 2/3 % more than that in bag ‘B’ and probability of drawing a black cap from bag ‘B’ is (3/7).

Correct Answer: (d) 4  
Solution:

Total number of caps in bag ‘A’
= x − 4 + 9 + x − 1 = 2x + 4

ATQ:
{2(x − 4)(x − 1)}/[(2x + 4)(2x + 3)] = 2/7

7(x − 4)(x − 1) = (2x + 4)(2x + 3)
7(x² − 5x + 4) = (4x² + 14x + 12)
7x² − 35x + 28 = 4x² + 14x + 12
3x² − 49x + 16 = 0
3x² − 48x − x + 16 = 0
3x(x − 16) − 1(x − 16) = 0
(3x − 1)(x − 16) = 0
x = 16

Number of yellow caps in bag ‘B’ = 20 − (16 − 4) = 8
Number of black caps in bag ‘B’ = 3/5 × (16 − 1) = 9

Let number of white caps in bag ‘B’ be ‘y’
So, 9/(9 + 8 + y) = 3/7
y + 17 = 21
y = 4

Hence, option d.

4. Directions (4-6): Answer the questions based on the information given below.

There are two friends A and B, playing different games consisting of dices and cards. There are three types of game i.e. game 1, game 2 and game 3. In game 1, both of them are allowed to throw a dice alternately. If 6 appear on throwing, the person is said to be the winner. In game 2, they throw a pair of dice alternately. A wins if he gets a sum of 8 before B gets a sum of 9 and B wins if he gets a sum of 9 before A gets a sum of 8. In game 3, both of them are allowed to throw a card from a well shuffled deck of 52 cards without replacement. Any person is said to win this game, if he throws a face card.

Ques: If B begins the game 1, find the probability of B’s winning in his third attempt.

Correct Answer: (d) 625/7776  
Solution:

Probability of getting a six = 1/6
Probability of not getting a six = 1 − 1/6 = 5/6

Since, B starts the game.
So, the probability of B’s winning in his third attempt
= 5/6 × 5/6 × 5/6 × 5/6 × 1/6
= (5/6)⁴ × 1/6
= 625/7776

Hence, option d

5. If A begins the game 2, then find the probability of B’s winning.

Correct Answer: (c) 31/76
Solution:

A pair of dice is thrown.
Total sample space = 6² = 36

So, 8 can come in the following ways,
(2,6), (6,2), (3,5), (5,3), and (4,4).
So, probability of throwing 8 = 5/36

Probability of not throwing 8 = 1 − (5/36)
= 31/36

And 9 can come in the following ways,
(3,6), (6,3), (4,5) and (5,4).
So, probability of throwing 9 = 4/36 = 1/9

Probability of not throwing 9 = 1 − (1/9) = 8/9

A can win in first, third, fifth, seventh….. throws

So, probability of A’s winning in first throw = 5/36

Probability of A’s winning in third throw = 31/36 × 8/9 × 5/36

Probability of A’s winning in fifth throw
= (31/36)² × (8/9)² × 5/36

Hence, probability of A’s winning
= 5/36 + 31/36 × 8/9 × 5/36 + (31/36)² × (8/9)² × 5/36 + …….

= (5/36) / (1 − (31/36 × 8/9))
= 45/76

Therefore, probability of B’s winning
= 1 − probability of A’s winning
= 1 − 45/76 = 31/76

Hence, option c.

6. If A starts the game 3, find the probability of B’s winning in his second attempt

Correct Answer: (c) 456/4165
Solution:

Total number of face cards in a deck of 52 cards = 12
Required probability = 40/52 × 39/51 × 38/50 × 12/49
= 456/4165

Hence, option c.

7. Directions (7-8): Answer the questions based on the information given below.

A box contains some red and some green bottles. The difference between the number of red and green bottles is 5. The probability of choosing two bottles of different colour is equal to the probability of choosing two bottles of same colour.

Ques: What is the total number of bottles in the box?

Correct Answer: (c) 25
Solution:

Let the number of red bottles = x
The number of green bottles = (x + 5) or (x - 5)
Case1:
If the number of red bottles and green bottles be ‘x’
and ‘(x + 5)’, respectively.
According to the question,
Probability of choosing two bottles of different colour
= [(ˣC₁ × ⁽ˣ⁺⁵⁾C₁)/(²ˣ⁺⁵C₂)]
= [2x × (x + 5)]/[(2x + 5)(2x + 5 - 1)]
Probability of choosing two bottles of same colour
= [(ˣC₂/(²ˣ⁺⁵C₂)) + (⁽ˣ⁺⁵⁾C₂/(²ˣ⁺⁵C₂)]
= [x(x - 1)]/[(2x + 5)(2x + 5 - 1)] + [(x + 5)(x + 5-1)]/[(2x + 5)(2x + 5 - 1)]
According to the question,
[2x × (x + 5)]/[(2x + 5)(2x + 5 - 1)] = [x(x - 1)]/[(2x5)(2x + 5 - 1)] + [(x + 5)(x + 5 - 1)]/[(2x + 5)(2x+5 - 1)]
[2x × (x + 5)]/[(2x + 5)(2x + 5 - 1)] = [x(x - 1) + (x+5)(x + 5 - 1)]/[(2x + 5)(2x + 5 - 1)]
2x² + 10x = x² - x + (x + 5)(x + 4)
2x² + 10x = x² - x + x² + 9x + 20
10x = 8x + 20
2x = 20
x = 10
So, the number of red bottles = 10
The number of green bottles = 15
Case 2:
If the number of red bottles and green bottles be ‘x’
and ‘(x - 5)’, respectively
According to the question,
Probability of choosing two bottles of different colour
= [(ˣC₁ × ⁽ˣ⁻⁵⁾C₁)/(²ˣ⁻⁵C₂)]
= [2x × (x - 5)]/[(2x - 5)(2x - 5 - 1)]
Probability of choosing two bottles of same colour
= [(ˣC₂/(²ˣ⁻⁵C₂)) + (⁽ˣ⁻⁵⁾C₂/(²ˣ⁻⁵C₂))]
= [x(x - 1)]/[(2x - 5)(2x - 5 - 1)] + [(x - 5)(x - 5 -1)]/[(2x - 5)(2x - 5 - 1)]
According to the question,
[2x × (x - 5)]/[(2x - 5)(2x - 5 - 1)] = [x(x - 1)]/
[(2x - 5)(2x - 5 - 1)] + [(x - 5)(x - 5 - 1)]/[(2x - 5)(2x-5 - 1)]
[2x × (x - 5)]/[(2x - 5)(2x - 5 - 1)] = [x(x - 1) +
(x - 5)(x - 6)]/[(2x - 5)(2x - 5 - 1)]
2x² - 10x = x² - x + (x - 5)(x - 6)
2x² - 10x = x² - x + x² - 11x + 30-10x = 30 - 12x
2x = 30
x = 15
So, the number of red bottles = 15
The number of green bottles = 10
The total number of bottles in the box = 10 + 15 = 25
Hence, option c.

8. If the probability of choosing two red bottles from the box is 3/20, then find the number of red bottles in the box

Correct Answer: (b) 10  
Solution:Let the number of red bottles = x
The number of green bottles = (x + 5) or (x − 5)

Case1:
If the number of red bottles and green bottles be ‘x’
and ‘(x + 5)’, respectively.

According to the question,
Probability of choosing two bottles of different colour
= [{xC₁ × (x+5)C₁}/{(2x+5)C₂}]
= [2x × (x + 5)]/[(2x + 5)(2x + 5 − 1)]

Probability of choosing two bottles of same colour
= [(xC₂/(2x+5)C₂) + ((x+5)C₂/(2x+5)C₂)]
= [x(x − 1)]/[(2x + 5)(2x + 5 − 1)] + [(x + 5)(x + 5 − 1)]/[(2x + 5)(2x + 5 − 1)]

According to the question,
[2x × (x + 5)]/[(2x + 5)(2x + 5 − 1)] = [x(x − 1)]/[(2x + 5)(2x + 5 − 1)] + [(x + 5)(x + 5 − 1)]/[(2x + 5)(2x + 5 − 1)]

[2x × (x + 5)]/[(2x + 5)(2x + 5 − 1)] = [x(x − 1) + (x + 5)(x + 5 − 1)]/[(2x + 5)(2x + 5 − 1)]

2x² + 10x = x² − x + (x + 5)(x + 4)
2x² + 10x = x² − x + x² + 9x + 20
10x = 8x + 20
2x = 20
x = 10

So, the number of red bottles = 10
The number of green bottles = 15

Case 2:
If the number of red bottles and green bottles be ‘x’
and ‘(x − 5)’, respectively.

According to the question,
Probability of choosing two bottles of different colour
= [{xC₁ × (x−5)C₁}/{(2x−5)C₂}]
= [2x × (x − 5)]/[(2x − 5)(2x − 5 − 1)]

Probability of choosing two bottles of same colour
= [(xC₂/(2x−5)C₂) + ((x−5)C₂/(2x−5)C₂)]
= [x(x − 1)]/[(2x − 5)(2x − 5 − 1)] + [(x − 5)(x − 5 − 1)]/[(2x − 5)(2x − 5 − 1)]

According to the question,
[2x × (x − 5)]/[(2x − 5)(2x − 5 − 1)] = [x(x − 1)]/[(2x − 5)(2x − 5 − 1)] + [(x − 5)(x − 5 − 1)]/[(2x − 5)(2x − 5 − 1)]

[2x × (x − 5)]/[(2x − 5)(2x − 5 − 1)] = [x(x − 1) + (x − 5)(x − 6)]/[(2x − 5)(2x − 5 − 1)]

2x² − 10x = x² − x + (x − 5)(x − 6)
2x² − 10x = x² − x + x² − 11x + 30
−10x = 30 − 12x
2x = 30
x = 15

So, the number of red bottles = 15
The number of green bottles = 10

Case 1:
If the number of red bottles in the box = 15
Probability of choosing 2 red bottles = (15C₂/25C₂)
= (15 × 14)/(25 × 24) = 7/10
This is not satisfying the condition.

Case 2:
If the number of red bottles in the box = 10
Probability of choosing 2 red bottles = (10C₂/25C₂)
= (10 × 9)/(25 × 24) = 3/20
This is satisfying the condition.

Hence, option b.

9. Directions (9-10): Answer the questions based on the information given below.

Bag ‘A’: There are (x + 2) red gems, (x - 4) blue gems and 7 green gems. The probability of picking a red gem is 1/5 more than the probability of picking a green gem.
Bag ‘B’: Number of red gem is 25% more than that of number of red gem in bag ‘A’. There are (x - 8) blue gems and ‘y’ green gems such that probability of picking two green gems at random is 10/231. Total number of gems in the bag is less than 24.

Ques: Find the value of (x - y).

Correct Answer: (c) 5
Solution:

Total number of gems in bag ‘A’ = (x + 2 + x − 4 + 7)
= (2x + 5)

According to the question,
{[(x + 2)/(2x + 5)] − [7/(2x + 5)]} = 1/5
5 × (x + 2 − 7) = 2x + 5
5x − 25 = 2x + 5
3x = 30
x = 10

Therefore, in bag ‘A’:
Number of red gems = x + 2 = 12
Number of blue gems = x − 4 = 6
Number of green gems = 7

In Bag ‘B’:
Number of red gems = 1.25 × 12 = 15
Number of blue gems = (x − 8) = 2
Number of green gems = ‘y’

Total number of gems in the bag = (y + 2 + 15)
= (y + 17)

According to the question,
(yC₂/(y + 17)C₂) = 10/231

y(y − 1)/[(y + 17)(y + 16)] = 10/231
231y² − 231y = 10(y² + 33y + 272)
221y² − 561y − 2720 = 0

Now, total number of gems is less than 24, therefore,
‘y’ can be 6, 5, 4, 3, 2, 1

Also, in numerator of the probability, there is a 10, therefore ‘y’ should be 5 or 6
But, 6 will not satisfy the equation

Therefore,
221y² − 1105y + 544y − 2720 = 0
y(y − 5) + 544(y − 5) = 0
(y − 5)(y + 544) = 0
y = 5 (Because number of gems cannot be negative)

Required value = 10 − 5 = 5
Hence, option c.

10. Find the difference between probability of picking a blue gem from bag ‘A’ and a red gem from bag ‘B’.

Correct Answer: (e) 243/550
Solution:

Total number of gems in bag ‘A’ = (x + 2 + x − 4 + 7)
= (2x + 5)

According to the question,
{[(x + 2)/(2x + 5)] − [7/(2x + 5)]} = 1/5

5 × (x + 2 − 7) = 2x + 5
5x − 25 = 2x + 5
3x = 30
x = 10

Therefore, in bag ‘A’:
Number of red gems = x + 2 = 12
Number of blue gems = x − 4 = 6
Number of green gems = 7

In Bag ‘B’:
Number of red gems = 1.25 × 12 = 15
Number of blue gems = (x − 8) = 2
Number of green gems = ‘y’

Total number of gems in the bag = (y + 2 + 15)
= (y + 17)

According to the question,
(yC₂/(y + 17)C₂) = 10/231

y(y − 1)/[(y + 17)(y + 16)] = 10/231
231y² − 231y = 10(y² + 33y + 272)
221y² − 561y − 2720 = 0

Now, total number of gems is less than 24, therefore,
‘y’ can be 6, 5, 4, 3, 2, 1
Also, in numerator of the probability, there is a 10,
therefore ‘y’ should be 5 or 6
But, 6 will not satisfy the equation.
Therefore, 221y² − 1105y + 544y − 2720 = 0
y(y − 5) + 544(y − 5) = 0
(y − 5)(y + 544) = 0
y = 5 (Because number of gems cannot be negative)

Required difference = (15/22) − (6/25) = [(375 − 132)/550] = 243/550
Hence, option e.