First slide

Theorems of probability

Question

$L e t A d e n o t e t h e e v e n t t h a t a 6 - d i g i t i n t e g e r f o r m e d b y 0, 1, 2, 3, 4, 5, 6 w i t h o u t r e p e t i t i o n s, b e d i v i s i b l e b y 3 . T h e n p r o b a b i l i t y o f e v e n t A i s e q u a l t o :$

Moderate

A

$\frac{11}{27}$
B

$\frac{4}{9}$
C

$\frac{9}{56}$
D

$\frac{3}{7}$

Solution

$T o t a l n u m b e r o f 6 d i g i t e d n u m b e r s f r o m 0, 1, 2, 3, 4, 5, 6$
$= 6 \times 6 \times 5 \times 4 \times 3 \times 2 ∴ n (S) = 6 \times 6!; W e h a v e 0 + 1 + 2 + 3 + 4 + 5 + 6 = 21$
$T h e r e f o r e 6 d i g i t e d n u m b e r d i v i s i b l e b y 3 c a n b e f o r m e d b y u \sin g t h e d i g i t s 1, 2, 3, 4, 5, 6 i n 6! w a y s . B y u \sin g t h e d i g i t s 0, 1, 2, 4, 5, 6 i n 5 \times 5! w a y s$
$by using the digits 0, 1, 2, 3, 4, 5 in 5 \times 5! ways H e n c e, P (E) = \frac{6! + 2 \times 5 \times 5!}{6 \times 6!} = \frac{6 + 10}{36} = \frac{4}{9}$

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