First slide

Theorems of probability

Question

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

Easy

A

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

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

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

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

Solution

$T h e n u m b e r o f w a y s o f a r r a i n g i n g 7 b o y s a m o n g t h e m s e l v e s i s n (S) = 7! L e t E b e t h e e v e n t o f h a v i n g t h r e e p e r s o n s A, B, C a r e t o g e t h e r i n a p e r t i c u l a r o r d e r$
$T a k e A, B, C a s o n e b u n c h, t h e n u m b e r o f w a y s o f a r r a n g i n g 5 t h i n g s i n 5! w a y s$
$n (E) = 5!$
$T h e r e f o r e, t h e p r o b a b i l i t y i s P (E) = \frac{5!}{7!}$

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