First slide

Theorems of probability

Question

Seven persons sit in a row at random. The probability that the three persons A, B, C sit together is

Easy

A

$\frac{3!}{7!}$
B

$\frac{4!}{7!}$
C

$\frac{5!}{7!}$
D

$\frac{3! 5!}{7!}$

Solution

$n (S) = 7!$
$n (A) = 5! 3!$
$P (A) = \frac{5! 3!}{7!} = \frac{1}{7}$

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