BANK & INSURANCE (PROBABILITY) PART 2

Total Questions: 45

11. A rack has 3 green shirts, 4 blue shirts and 2 red shirts. If 2 shirts are picked at random without replacement from the rack, then find the probability that both shirts are of the same colour.

Correct Answer: (d) (5/18)  
Solution:

Probability of picking 2 green shirts = ³C₂ ÷ ⁹C₂
= 3/36 = (1/12)
Probability of picking 2 blue shirts = ⁴C₂ ÷ ⁹C₂
= 6/36 = (2/12)
Probability of picking 2 red shirts = ²C₂ ÷ ⁹C₂
= (1/36)
So, required probability = (3/36) + (6/36) + (1/36)
= (10/36) = (5/18)

12. The probability of Ram failing in his math test is (3/8) and that of him failing in his science test is (2/5). What is the probability of Ram failing in exactly one of the given two tests?

Correct Answer: (b) (19/40)  
Solution:Probability of passing in maths test = 1 − (3/8) = (5/8)
Probability of passing in science test = 1 − (2/5) = (3/5)
Required probability = Probability of failing in maths test × probability of passing in science test + probability of passing in maths test × probability of failing in science test
= (3/8) × (3/5) + (5/8) × (2/5) = (9/40) + (10/40) = (19/40)

13. Out of Idly, Dosa and Poha, Shivang can eat only one item. The probability that Shivang eats Idly is 0.12 more than the probability that he eats Dosa or Poha. If the probability that he eats Poha is (3/25), then find the probability that he eats Dosa

Correct Answer: (d) 0.32  
Solution:Let the probability that Shivang eats Dosa or poha be ‘x’
Then, probability that Shivang eats Idly = (x + 0.12)
So, x + x + 0.12 = 1
2x + 0.12 = 1
So, x = (1 − 0.12) ÷ 2 = 0.44
So, probability that he eats Dosa = 0.44 − (3/25)
= {(11 − 3)/25} = 8/25 = 0.32
Hence, option d.

14. On a biased dice, an even number appears three times as frequently as an odd number. If dice is thrown twice, what is the probability that the sum is 10 or more?

Correct Answer: (b) 17/72  
Solution:

Let the probability of getting an odd number (1, 3 or 5) on a dice be x, then the probability of getting an even number (2, 4 or 6) would be 3x. Now we have 3 odd numbers and 3 even numbers, the sum of the probability should be 1, so x + x + x + 3x + 3x + 3x = 1, so 12x = 1, x = 1/12, therefore, probability of an odd number = 1/12 and probability of an even number = 1/4.

Probability (sum is 10 or more) = P (of getting sum 10) + P (of getting sum 11) + P (of getting sum 12)
P (of getting sum 10) = P(4) × P (6) + P (5) × P (5) + P (6) × P (4)
= 1/4 × 1/4 + 1/12 × 1/12 + 1/4 × 1/4 = 19/144
P (of getting sum 11) = P (5) × P (6) + P (6) × P (5)
= 1/12 × 1/4 + 1/4 × 1/12 = 1/24
P (of getting sum 12) = P (6) × P (6) = 1/4 × 1/4 = 1/16

Required probability = 19/144 + 1/24 + 1/16
= 34/144 or 17/72.

15. A bag contains 6 black and 2 white balls and another bag contains 6 black and 8 white balls. If one of the bag is selected at random and two balls are selected at random from the bag thus selected, then what is the probability that the two balls selected are of different colours?

Correct Answer: (e) None of these
Solution:Probability of selecting any bag = 1/2
Probability of selecting a black and a white ball from
1ˢᵗ bag = (⁶C₁ × ²C₁)/⁸C₂ = 3/7
Probability of selecting a black and a white ball from
2ⁿᵈ bag (⁶C₁ × ⁸C₁)/¹⁴C₂ = 24/91
Required probability = P (1ˢᵗ bag) P (a black ball and
a white ball from, 1ˢᵗ bag) + P (2ⁿᵈ bag) P (a black ball
and a white ball from 2ⁿᵈ bag)
= 1/2 × 3/7 + 1/2 × 24/91 = 63/182

16. In a test, the probability of Ajay scoring a 100/100 is 1/5. In a university examination of 6 papers, what is the probability (in approximate percentage) of Ajay scoring a 100/100 in at least 2 papers?

Correct Answer: (b) 34%  
Solution:

Probability that Ajay scores 100/100 in exactly 2 papers = ⁶C₂ × (1/5)² (4/5)⁴ = 0.24576
Probability that Ajay scores 100/100 in exactly 3 papers = ⁶C₃ × (1/5)³ (4/5)³ = 0.08192
Probability that Ajay scores 100/100 in exactly 4 papers = ⁶C₄ × (1/5)⁴ (4/5)² = 0.01536
Probability that Ajay scores 100/100 in exactly 5 papers = ⁶C₅ × (1/5)⁵ (4/5) = 0.001536
Probability that Ajay scores 100/100 in exactly 6 papers = ⁶C₆ × (1/5)⁶ = 0.000064

Required probability = 0.24576 + 0.08192 + 0.01536 + 0.001536 + 0.000064 = 0.34464 ~ 34%.

17. There are two bags, one of which contains 6 black and 8 white balls, while the other contains 8 black and 6 white balls. A dice is cast. If the face 2 or 6 turns up, a ball is taken from the first bag and if any other face turns up a ball is chosen from the second bag. The probability of choosing a black ball is

Correct Answer: (b) 11/21  
Solution:Bag 1 contains 6 black and 8 white balls.
Bag 2 contains 8 black and 6 white balls.
P (Bag 1 is chosen) = 2/6 = 1/3
P (Bag 2 is chosen) = 4/6 = 2/3
P (black ball is drawn from bag 1) = 6/14 = 3/7
P (black ball is drawn from bag 2) = 8/14 = 4/7
P (black ball) = 3/7 × 1/3 + 4/7 × 2/3 = 1/7 + 8/21 = 11/21

18. A bag contains 3 blue and 4 black balls and another bag contains 4 blue and 3 black balls. A dice is cast and if the face 2 or 5 turns up, a ball is taken from the first bag and if any other face turns up, a ball is taken from the second bag. What would have been the increase/decrease in probability of drawing a black ball if the first bag was selected when the dice’s face turned out to be 1,2 or 5, and bag 2 was selected when it showed any other number.

Correct Answer: (c) 1/42 increase  
Solution:The probability of selection of first bag = 1/3.
The probability of selecting a black ball from the first bag = 4/7.
The probability of selection of first bag = 2/3.
Probability of selecting a black ball from the second bag = 3/7.
Probability of selecting a black ball = 4/21 + 6/21 = 10/21.
In the second case, probability of selecting both bags become 1/2.
Therefore, change = 4/14 + 3/14 − 10/21 = 1/42 increase.

19. Urn A contains six blue and four black balls and urn B contains four blue and six black balls. One ball is drawn at random from urn A and placed in urn B. Then one ball is transferred at random from urn B to urn A. If one ball is now drawn at random from urn A, find the probability that it is blue.

Correct Answer: (a) 32/55  
Solution:

The probability that a blue ball is transferred from A to B and a blue ball is transferred from urn B to urn A,
Pblue,blue = (6/10) × (5/11) = 30/110.
Similarly other cases, Pblue,black = (6/10) × (6/11) = 36/110.
Pblack,blue = (4/10) × (4/11) = 16/110.
Pblack,black = (4/10) × (7/11) = 28/110.

Probability of drawing a blue ball from urn A after transfers in these four cases are 6/10, 5/10, 7/10, 6/10 respectively.
Therefore the required probability = (30/110) × (6/10) + (5/10) × (36/110) + (7/10) × (16/110) + (6/10) × (28/110) = 32/55

20. Raj and Simran alternately throwing a pair of dice. Raj wins if he throws 5 before Simran throws 6 and Simran wins if she throws 6 before Raj throws 5. Find their respective chances of winning, if Raj begins.

Correct Answer: (b) 9/19, 10/19
Solution:

5 can be thrown with a pair of dice in the following ways: (1,4), (2,3), (3,2), (4,1).
Probability of throwing a 5 = 1/9.
Probability of not throwing a 5 = 8/9.

6 can be thrown in the following ways: (1,5), (2,4), (3,3), (4,2), (5,1).
Probability of throwing 6 = 5/36
Probability of not throwing a 6 = 31/36.

Raj wins if he throws 5 in the first, third and fifth throws. And Simran is not able to throw 6 in second, fourth throws …
Thus probability of Raj winning
= 1/9 + (8/9) × (31/36) × (1/9) + (8/9) × (31/36) × (8/9) × (31/36) × (1/9) + …
= (1/9) × (1 − (8/9 × 31/36)) = 9/19.
Probability of winning of Simran = 10/19