Accenture Practice Probability Questions
In Accenture papers, you can expect few questions from probability. These would generally of conventional nature and hence you can answer these with good practice.
Question 1
A leather box contains 8 black balls and 6 white balls. Two draws of three balls each are made, the balls being replaced after the first draw. What is the probability that the balls were black in the first draw and white in the second draw?
a) 70/8281 b)140/20449 c ) 25/5445 d)35/5448
Answer : a)70/8281
Solution:
Total number of balls = 8 + 6 = 14
Total ways of drawing 3 balls, N(S) = 14C3 = 364
No of ways to draw 3 black balls = N(E1) for black balls = 8C3 = 56
Probability of all balls being black = P(E1) = N(E1) / N(S) = 56/ 364 = 14/91
No of ways to draw 3 white balls = N(E2) for white balls = 6C3 = 20
Probability of all balls being white = P(E2) = 20 / 364 = 5/91
Probability of drawing 3 black balls in first draw and drawing 3 white balls in the second draw = P(E1) x P(E2)
Therefore P(E) = 14/91 x 5/91 = 70 /8281
Question 2
Fourteen persons are sitting around a circular table facing the centre. What is the probability that three particular persons sit together?
a) 2/9 b)1/13 c)2/13 d)1/26
Answer : d) 1/26
Solution :
In a circle of n different persons, the total number of arrangements possible = (n - 1)!
Total number of arrangements = n(S) = (14 – 1)! = 13 !
Taking three persons as a unit, total persons = 12 (in 4 units)
Therefore no. of ways for these 12 persons to around the circular table = (12 - 1)! = 11!
In any given unit, 3 particular person can sit in 3!. Hence total number of ways that any three person can sit =
n(E) = 11! X 3!
Therefore P (E) = probability of three persons sitting together = n(E) / n(S) = 11! X 3 ! divided by 13!
3 x 2 divided by 13 x 12 = 1/26
Question 3
In how many different ways can the letter of the word “ECHRONICLEE” be arranged?
a)1663200 b)8316000 c)3326400 d)4158000
Answer : c)3326400
Solution :
ECHRONICLEE has 11 letters
Total number of rearrangements = (Number of letters)! / (Number of 1st repetitive letter)! x (Number of 2nd repetitive letter)!....
In the word ECHRONICLEE, E occurs thrice. C occurs twice. Hence denominator of the above formula will become 3! x 2!
Therefore Total number of rearrangements = 11!/ 3! 2!
= 3326400




