Below are three probability based questions to practice for Accenture papers.
Question 1
A box contains 4 blue and 5 white balls and another box contains 5 blue and 4 white balls. One ball is to be drawn from either of the two boxes. What is the probability of drawing a blue ball?
a) 2/9 b) 1/9 c) 4/9 d) 1/2
Answer : d) 1/2
Solution :
Probability of choosing first box = 1/2 and
Probability of choosing second box = 1/2
Imagine there is no second box at all :
P1 = Probability of choosing one blue ball from first box when there is no second box at all
= No of ways of choosing 1 blue ball from among 4 blue balls / Total number of ways of choosing 1 ball from all 9 balls.
= 4C1 / 9C1
Imagine there is no first box at all :
P2 = Probability of choosing one blue ball from second box when there is no first box at all
= No of ways of choosing 1 blue ball from among 5 blue balls / Total number of ways of choosing 1 ball from all 9 balls.
= 5C1 / 9C1
Now, consider our real scenario of having both the boxes :
P3 = Probability of choosing one blue ball from the first box = P1 x Probability of choosing first box = 4C1 / 9C1 x 1 / 2 = 2/9
P4 = Similarly probability of choosing one blue ball from the second box = P2 x Probability of choosing second box = 5C1 / 9C1 x 1 / 2 = 5/18
P (E) = Probability of choosing a blue ball from either of the boxes =
P3 + P4 = 2/9 + 5/18 = 1/2 --> eq 1
Note : Since the question reads we have to choose one blue ball from first OR second box we have used '+' sign in eq 1. If the question were choosing one blue ball from first AND second box, we should had used multiplication i.e ' x '.
Question 2
An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If four marbles are picked at random, what is the probability that at least one is blue?
a) 4/15 b) 69/91 c) 11/15 d) 22/91
Answer : b) 69/91 .
Solution :
Total possible outcomes = n(S)
= Selection of 4 marbles out of 15 marbles
= 15C4 = (15 x 14 x 13 x 12) divided by 1 x 2 x 3 x4 = 1365
When no marble is blue, favourable number of cases n(E)
= Selection of 4 marbles out of 11 marbles
= 11C4 = 11 x 10 x 9 x 8 divided by 1 x 2 x 3 x 4 = 330
Probability that atleast one ball is blue + Probability that no ball is blue = 1
Therefore, Probability that atleast one ball is blue = 1 - Probability that no ball is blue
Therefore, Probability that atleast one ball is blue = 1 - n(E) / n(S)
= 1 - 330/1365 = 1 - 22/91 = 69 /91
Question 3
A committee of 4 members is to be selected from a group of 4 women and 3 men. What is the probability that the committee has at least one man.?
a) 1/35 b) 3/7 c) 34/35 d) 4/7
Answer : c) 34/35
Solution :
Probability of there being all women in the committee = 4C4 divided by 7C4 = 1/35
But, Probability of there being all women in the committee + Probability of there being at least one man = Probability of formation of a committee with no restriction of gender = 1
Probability of there being at least one man = 1 - 1/35 = 34/35
Question 1
A box contains 4 blue and 5 white balls and another box contains 5 blue and 4 white balls. One ball is to be drawn from either of the two boxes. What is the probability of drawing a blue ball?
a) 2/9 b) 1/9 c) 4/9 d) 1/2
Answer : d) 1/2
Solution :
Probability of choosing first box = 1/2 and
Probability of choosing second box = 1/2
Imagine there is no second box at all :
P1 = Probability of choosing one blue ball from first box when there is no second box at all
= No of ways of choosing 1 blue ball from among 4 blue balls / Total number of ways of choosing 1 ball from all 9 balls.
= 4C1 / 9C1
Imagine there is no first box at all :
P2 = Probability of choosing one blue ball from second box when there is no first box at all
= No of ways of choosing 1 blue ball from among 5 blue balls / Total number of ways of choosing 1 ball from all 9 balls.
= 5C1 / 9C1
Now, consider our real scenario of having both the boxes :
P3 = Probability of choosing one blue ball from the first box = P1 x Probability of choosing first box = 4C1 / 9C1 x 1 / 2 = 2/9
P4 = Similarly probability of choosing one blue ball from the second box = P2 x Probability of choosing second box = 5C1 / 9C1 x 1 / 2 = 5/18
P (E) = Probability of choosing a blue ball from either of the boxes =
P3 + P4 = 2/9 + 5/18 = 1/2 --> eq 1
Note : Since the question reads we have to choose one blue ball from first OR second box we have used '+' sign in eq 1. If the question were choosing one blue ball from first AND second box, we should had used multiplication i.e ' x '.
Question 2
An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If four marbles are picked at random, what is the probability that at least one is blue?
a) 4/15 b) 69/91 c) 11/15 d) 22/91
Answer : b) 69/91 .
Solution :
Total possible outcomes = n(S)
= Selection of 4 marbles out of 15 marbles
= 15C4 = (15 x 14 x 13 x 12) divided by 1 x 2 x 3 x4 = 1365
When no marble is blue, favourable number of cases n(E)
= Selection of 4 marbles out of 11 marbles
= 11C4 = 11 x 10 x 9 x 8 divided by 1 x 2 x 3 x 4 = 330
Probability that atleast one ball is blue + Probability that no ball is blue = 1
Therefore, Probability that atleast one ball is blue = 1 - Probability that no ball is blue
Therefore, Probability that atleast one ball is blue = 1 - n(E) / n(S)
= 1 - 330/1365 = 1 - 22/91 = 69 /91
Question 3
A committee of 4 members is to be selected from a group of 4 women and 3 men. What is the probability that the committee has at least one man.?
a) 1/35 b) 3/7 c) 34/35 d) 4/7
Answer : c) 34/35
Solution :
Probability of there being all women in the committee = 4C4 divided by 7C4 = 1/35
But, Probability of there being all women in the committee + Probability of there being at least one man = Probability of formation of a committee with no restriction of gender = 1
Probability of there being at least one man = 1 - 1/35 = 34/35
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