Probability Question 118
Question: Ten students are seated at random in a row. The probability that two particular students are not seated side by side is
Options:
A) $ \frac{4}{5} $
B) $ \frac{3}{5} $
C) $ \frac{2}{5} $
D) $ \frac{1}{5} $
Show Answer
Answer:
Correct Answer: A
Solution:
Total ways $ =10! $
Two boys can sit side by side in $ 2\times 9! $ ways.
So probaibility $ =\frac{2\times 9!}{10!}=\frac{1}{5} $
Thus the probability that they are not seated together is $ 1-\frac{1}{5}=\frac{4}{5}. $
BETA
