Probability Question 243
Question: There are only two women among 20 persons taking part in a pleasure trip. The 20 persons are divided into two groups, each group consisting of 10 persons. Then the probability that the two women will be in the same group is
Options:
A) 9/19
B) 9/38
C) 9/35
D) none
Show Answer
Answer:
Correct Answer: A
Solution:
The number of ways in which 20 people can be divided into two equal groups is $ n(s)=\frac{20!}{10!10!2} $ The number of ways in which 18 people can be divided into groups of 10 and 8 is $ n(A)=\frac{18!}{10!8!} $ $ \therefore P(E)=\frac{18!}{10!8!}\frac{10!10!2}{20!}=\frac{10\times 9\times 2}{20\times 19}=\frac{9}{19} $
BETA
