Answer to Question #154343 in Statistics and Probability for Patel

Question

Eight boys and three girls are to sit in a row for a photograph. Find the probability that no two girls are together.

Answer 1

Given: $8$ boys; $3$ girls;

total people: $11$

total ways of arranging: $11!$

We can place the boys in 8! $= P^8_8$ ways and the girls can't sit next to each other,

there are only 9 seats for the girls, but there are only 3 girls hence there are only $P^9_3$ ways for them to sit.

All posible ways of arranging which are suitable for us: $P^8_8*P^9_3 = \frac{8!}{0!}*\frac{9!}{(9-3)!}, 0! =1$

$p =\frac{P^8_8*P^9_3}{11!} = \frac{8!}{0!}*\frac{9!}{(9-3)!11!} \approx 0.509091$

$p \approx 0.509091$