21-10-2021 Answers

 1. A bomber drops 4 bombs to destroy a target. Two bombs are enough to destroy the target. If the chance of a bomb hitting the target is 0.4, find the probability that the target is not destroyed.

Answer: 297/625

Solution: Probability of a bomb hitting the target = 0.4 = 2/5, not hitting = 3/5

The target is not destroyed only when just one or no bomb hits it.

Probability that all bombs miss = (3/5)^4 = 81/625

Probability that just one bomb hits = 4C1 * (2/5) * (3/5)^3 = 216/625

Probability that the target is NOT destroyed = 81/625 + 216/625 = 297/625

2. A classroom has 3 electric bulb holders. From a collection of 10 bulbs of which 4 are defective, 3 are selected at random and put in the holders. Find the probability that all bulbs are glowing.

Answer: 1/6

Solution: Number of ways to pick all three bulbs as working = 6C3 = 20

Total number of ways to pick 3 bulbs = 10C3 = 120

Hence probability that all bulbs are glowing = 20/120 = 1/6

3. In a company there are 5 engineers out of 20 workers. If 3 workers are selected at random what is the probability that all are engineers?

Answer: 1/114

Solution: Ways to pick 3 workers = 20C3 = 1140

Ways to pick 3 engineers from 5 = 5C3 = 10

Reqd probability that all are engineers = 10/1140 = 1/114

4. A bag contains 10 white and 15 black balls. Two balls are drawn in succession without replacement. What is the probability that first is white and second is black?

Answer: 1/4

Solution: Probability of drawing white as first ball = 10/25 = 2/5, black as second = 15/24 = 5/8

Reqd probability = 2/5 * 5/8 = 2/8 = 1/4

5. Find the probability that in a random arrangement of the letters of the word UNIVERSITY, the two I's are together.

Answer: 1/5

Solution: 10 letters are there in UNIVERSITY. Total number of unique words = 10!/2

Taking both I's together, there are effectively 9 letters. Total number of unique words with I's together = 9!

Hence required probability = 9!/(10!/2) = 2/10 = 1/5

6. A and B play a game where both are asked to select a number from 1 to 10. If both of them select the same number both of them receive a prize. Find the probability that they will not win a prize in a single trial.

Answer: 9/10

Solution: Assume A selects any number from 1 to 10.

Now B must not choose the number selected by A.That is B must select any of the remaining 9 numbers.

Hence the required probability = 9/10

7. A and B stand in a line at random with 10 other people. What is the probability that there are exactly 3 people between A and B?

Answer: 4/33

Solution: Total number of positions A and B can occupy = 12C2 = 66

Num of positions where 3 people can be between A and B = 12-3-1 = 8

Hence required probability = 8/66 = 4/33

8. Two letters are randomly picked from the word FOAM. Find the probability that at least one is a vowel

Answer: 5/6

Solution: Probability that no vowel is picked = 2C2/4C2 = 1/6

Hence probability that at least one vowel is picked = 1 - 1/6 = 5/6

9. The odds in favour of an event A are 3:4. The odds against another independent event B are 7:4. What is the probability that one of the events will happen?

Answer: 7/11

Solution: Probability of event A happening = 3/(3+4) = 3/7

Probability of event B NOT happening = 7/11. Hence B happening = 1-7/11 = 4/11.

As both are independent events, probability that one of the events will happen = 3/7 + 4/11 - (3/7 * 4/11)= (33+28-12)/77 = 49/77 = 7/11

10. 6 persons A,B,C,D,E,F are contesting for 3 seats. If D is elected uncontested, then the probability that C is elected is

Answer: 2/5

Solution: As D is already elected, remaining seats = 2, candidates = 5.

Num of ways to choose two candidates from five = 5C2 = 10

Num of ways to NOT to choose C = 4C2 = 6

Reqd probability that C is NOT elected = 6/10 = 3/5

Reqd probability that C is elected = 1 - 3/5 = 2/5

11. A box contains 4 red pens and 4 blue pens. Two pens are drawn at random. What is the probability that both are of same colour?

Answer: 3/7

Solution: Total ways of drawing two pens = 8C2 = 28

Ways of drawing two red pens = 4C2 = 6, blue pens = 4C2 =6

Required probability = (6+6)/28 = 12/28 = 3/7

12. Fivepersons A,B,C,D,E occupy five seats in a row randomly. What is the probability that A and B sit in adjacent seats?

Answer: 2/5

Solution: Total number of arrangements possible = 5! = 120

When A and B sit in adjacent seats, number of arrangements possible = 2 * 4! = 48 (We multiply by 2 because AB and BA are valid combinations)

Hence required probability = 48/120 = 2/5

13. In a simultaneous throw of two dice, find the probability of getting a total of 8.

Answer: 5/36

Solution: In a simultaneous throw of two dice, sample space = 6 * 6 = 36

Favourable cases = (2, 6) (3, 5) (4, 4) (5, 3) (6, 2)

So, the required probability = 5/36

14. A bag contains 8 red and 6 black balls. If 5 balls are drawn at random, what is the probability that 3 are red and 2 are black?

Answer: 60/143

Solution: Number of ways of drawing 5 balls from 14 balls = 14C5 = 2002

Number of ways of drawing 3 red and 2 black = 8C3 * 6C2 = 56*15 = 840

Required probability = 840/2002 = 420/1001 = 60/143

15. The odds in favour of an event A are 2:5. The odds against another independent event B are 5:1. What is the probability that one of the events will happen?

Answer: 17/42

Solution: Probability of event A happening = 2/7

Probability of event B NOT happening = 5/6. Hence B happening = 1-5/6 = 1/6.

As both are independent events, probability that one of the events will happen = 2/7 + 1/6 - (2/7 * 1/6)

= (12+7-2)/42 = 17/42