BANK & INSURANCE (PROBABILITY) PART 1

Total Questions: 45

11. A bag contains 13 white and 7 black balls. Two balls are drawn at random. What is the probability that they are of the same colour?

Correct Answer: (e) None of these
Solution:P (of getting two balls of same colour) = P (of getting two black balls) + P (of getting two white balls)
= ¹³C₂ / ²⁰C₂ + ⁷C₂ / ²⁰C₂ = 99/190

12. A bag contains 8 orange balls, 5 yellow balls, and 2 pink balls. If two balls are picked up at random, what is the probability that they are of same colour?

Correct Answer: (a) 13/35  
Solution:Total number of balls = 8 + 5 + 2 = 15
Total number of ways in which 2 balls can be selected are = ¹⁵C₂ ways = 105
2 orange balls are selected are = ⁸C₂ ways = 28
2 yellow balls = ⁵C₂ ways = 10
2 pink balls are selected = ²C₂ ways = 1
Required probability = (28 + 10 + 1)/105
= 13/35

13. The number of ways in which a cricket team of 11 players can be formed from 16 players always including the captain and always excluding 2 players is:

Correct Answer: (a) 286  
Solution:Captain always have to be selected and 2 players always have to be excluded, so we need to select 10 players from 13 which can be done in ¹³C₁₀ ways = 286 ways.

14. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

Correct Answer: (e) 25200
Solution:Out of 7 consonants 3 consonants is to be selected i.e. ⁷C₃ = 35
Out of 4 vowels 2 vowels is to be selected i.e. ⁴C₂ = 6
Total Number of ways = (35 × 6) = 210
Each group contains 5 letters and this can be arranges in 5! ways = 120
so, Required number of ways = (120 × 210)
= 25200

15. In a meeting everyone has shaken hands with everyone else; it was found that 210 handshakes were exchanged. How many members were present in the meeting?

Correct Answer: (c) 21
Solution:Let there be n persons in the meeting, then number of handshakes = ⁿC₂
ⁿC₂ = 210
n!/2!(n−2)! = 210
n(n−1) = 420
solving we get n = 21

16. A bag contains 20 coins of which 5 coins are counterfeit with tails on both sides. The rest are fair coins. One is selected at random from the bag and tossed. What is the probability of getting a head?

Correct Answer: (c) 3/8
Solution:There can be two possibilities either a counterfeit coin is selected or a fair coin is selected.
Probability of getting a head if the selected coin is counterfeit = 5/20 × 0 = 0
Probability of getting a head if the selected coin is fair = 15/20 × 1/2 = 3/8
Required probability = 0 + 3/8 = 3/8

17. A bag contains 3 red balls and 6 green balls. A second bag contains 7 red balls and 2 green balls. One bag is selected and then one ball is drawn from the selected bag. Find the probability that the ball is not red

Correct Answer: (a) 5/9  
Solution:

Probability of getting a red ball = 1/2 × 3/9 + 1/2 × 7/9 = 10/18 = 5/9
Probability of not getting a red ball = 1 − 5/9 = 4/9

18. Three dice are rolled simultaneously and the product of the three numbers appearing on the top is a prime number. What is the probability that the sum of the numbers appearing on the top is an odd number?

Correct Answer: (e) None of these
Solution:

Since product of three numbers appearing on the top of dice is a prime number, the possibilities are (1,1,2), (1,2,1), (2,1,1), (1,1,3) and (1,1,5). Among them (1,1,3) and (1,1,5) will give an odd sum.
No. of favourable ways = 3!/2! × 2 = 6
Total No. of ways = 3!/2! × 3 = 9
Required Probability = 6/9 = 2/3

19. From a box containing 20 bulbs, of which one-fifth are defective, three bulbs are chosen at random to fit into the three bulb holders in a room. What is the probability that the room is lighted?

Correct Answer: (e) None of these
Solution:

The room would be lighted if at least one non defective bulb is chosen. Required probability = 1 − (probability of all three being defective)
= 1 − (⁴C₃ / ²⁰C₃) = 1 − 4/1140 = 284/285

20. In a school, 35% of the students play chess, 55% play basketball and 10% both. If a student is selected at random, then the probability that he plays chess or basketball is:

Correct Answer: (e) None of these
Solution:Given that, 35% play chess; that is, P(C) = 35/100 = 7/20
55% play basketball; that is, P(B) = 55/100 = 11/20
And, 10% play both basketball and chess; that is, P(C ∩ B) = 10/100 = 1/10
Now, we have to find, the probability that 1 student plays chess or basketball; that we have to find, P(C ∪ B)
We know that, P(C ∪ B) = P(C) + P(B) − P(C ∩ B)
= 7/20 + 11/20 − 1/10 = 4/5