Probability 1 Question 21
29. Six boys and six girls sit in a row at random. Find the probability that
(i) the six girls sit together.
(ii) the boys and girls sit alternatively.
(1978, 3M)
Show Answer
Answer:
Correct Answer: 29. (i) $\frac{1}{132}$ (ii) $\frac{1}{462}$
Solution:
- (i) The total number of arrangements of six boys and six girls $=12$ !
$\therefore$ Required probability $=\frac{6 ! \times 7 !}{(12) !}=\frac{1}{132}$
[since, we consider six girls at one person]
(ii) Required probability $=\frac{2 \times 6 ! \times 6 !}{(12) !}=\frac{1}{462}$
