First slide

Theorems of probability

Question

Six faces of a die are marked with numbers 1, –1, 0, –2, 2, 3 and the die is thrown thrice. The probability that the sum of the numbers thrown is six, is

Easy

A

$\frac{3}{216}$
B

$\frac{6}{216}$
C

$\frac{10}{216}$
D

$\frac{18}{216}$

Solution

$T o t a l n u m b e r o f c a s e s i s n (S) = 6^{3} = 216$
Suppose that E be the event of getting sum 6
$n (E) = {(2, 2, 2), (3, 2, 1), (3, 1, 2), (1, 3, 2), (3, 3, 0), (3, 0, 3), (0, 3, 3), (1, 2, 3), (2, 1, 3), (2, 3, 1)}$
$∴ n (E) = 10 w a y s$
The probability that the sum of the numbers thrown is six, is $P (E) = \frac{10}{216}$

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