## Infosys Probability Questions and Solutions

1. There are 1000 junior and 800 senior students in a class.And there are 60 sibling pairs where each pair has 1 junior and 1 senior. One student is chosen from senior and 1 from junior randomly.What is the probability that the two selected students are from a sibling pair?

a.714 / 80000

b.724 / 80000

c.741 / 80000

d.722 / 70000

Ans: a

Sol:

Junior students = 1000

Senior students = 800

60 sibling pair = 2 x 60 = 120 student

One student chosen from senior = 800 C 1

=800

One student chosen from junior= 1000 C 1 =1000

Therefore, one student chosen from senior and one student chosen from junior n(s) = 800 x 1000=800000

Two selected students are from a sibling pair n(E)= 120 C 2 =7140

therefore,P(E) = n(E) / n(S)=7140/800000 = 714/80000

2. If a refrigerator contains 12 cans such that 7 blue cans and 5 red cans. In how many ways can we remove 8 cans so that at least 1 blue can and 1 red can remain in the refrigerator.

a.455

b.521

c.441

d.213

Ans: a

Sol:

Possible ways to draw 8 balls from the refrigerator which contains at least 1 blue and 1 red can after the drawing are (6,2) (5,3) (4,4).

For (6, 2) = ⇒ 7c6*5c2 ⇒ 7*10=70

For (5, 3) = ⇒ 7c5*5c3 ⇒ 21*10=210

For (4, 4) = ⇒ 7c4*5c4 ⇒ 35*5=175 So Total ways = 70+210+175=455

3.A Jar contains 18 balls. 3 blue balls are removed from the jar and not replaced.Now the probability of getting a blue ball is 1/5 then how many blue balls jar contains initially?

a.5

b.6

c.7

d.8

Ans: b

Sol:

x/15 = 1/5

x=3

3 + 3 (removed 3 blue balls) = 6

4.There is a school were 60% are girls and 35% of the girls are poor. Students are selected at random, what is the probability of selecting a poor girl out of total strength?

A.14/39

B.35/100

C.21/100

D.22/45

Ans. C

Explanation:

Let take 100 students outoff 100 60 are girls among girls poor girls are =35% 0f 60=21 pbt of selecting poor girl in total strength=21/100

5.What is the probability of drawn an ace or a space or both from a dew of cards.

A.13/52

B.26/52

C.16/52

D.16/50

Ans: C

Explanation:

There are 13 spades in a standard deck of cards. There are four aces in a standard deck of cards. One of the aces is a spade. So, 13 + 4 – 1 = 16 spades or aces to choose from. Since we have a total of 52 cards, the probability of selecting an ace or a spade is 16 / 52.

6. You are given three coins: one has heads on both faces, the second has tails on both faces, and the third has a head on one face and a tail on the other. You choose a coin at random and toss it, and it comes up heads. The probability that the other face is tails is

A.1/4

B.1/3

C.1/2

D.2/3

Ans: B

Explanation:

ans = 1/3 becoz ther is only one chance to get tail on other side of coin,amnog three coins,according to the question

7. In a horse racing competition, there were 18 numbered 1 to 18.The organizers assigned a probability of winning the race to each horse based on horses health and training the probability that horse one would win is 1/7, that 2 would win is 1/8, and that 3 would win is 1/7.Assuming that tie is impossible. Find the chance that one of these three will win the race?

A.22/392

B.1/392

C.23/56

D.391/392

Ans:C

Explanation:

CORRECT ANSWER IS 23/56 HORSE 1: 1/7 WINNING PROBABILITY HORSE 2: 1/8 WINNING PROBABILITY HORSE 3: 1/7 WINNING PROBABILITY ONE OF THESE WIN THE RACE: => 1/7 + 1/8 + 1/7 => 8/56 +7/56 + 8/56 (TAKING LCM) => 23/56

8.There are 1000 junior and 800 senior students in a class. And there are 60 sibling pairs where each pair has 1 junior and 1 senior.1 student is chosen from senior and 1 from junior randomly.What is the probability that the two selected students are from a sibling pair?

A.7140/800000

B.8450/800000

C.7455/800000

D.8230/800000

Ans:A

Explanation:

Junior student = 1000 Senior student = 800 60 sibling pair = 2 x 60 = 120 student Probability that 1 student chosen from senior = 800 Probability that 1 student chosen from junior = 1000 Therefore,1 student chosen from senior and 1 student chosen from junior n(s) = 800 x 1000 = 800000 Two selected student are from a sibling pair n(E) = 120C2 = 7140 Therefore P(E) = n(E)/n(S) = 7140⁄800000

9.There were two candidates in an election. Winner candidate received 62% of votes and won the election by 288 votes. Find the number of votes cast to the winning candidate?

A.456

B.744

C.912

D.1200

Ans:B

Explanation:

W = 62% L = 38% 62% – 38% = 24% 24% ——– 288 62% ——– ? => 744

10.A candidate who gets 30% of the marks fails by 50 marks. But another candidate who gets 45% marks gets 25 marks more than necessary for passing. Find the number of marks for passing?

A.150

B.200

C.250

D.275

Ans:B

Explanation:

30% ———— 50 45% ———— 25 ———————- 15% ————- 75 30% ————– ? 150 + 50 = 200 Marks