First slide

Theorems of probability

Question

Three dice are rolled and told that exactly two of them are showing the same number. The probability of getting sum 16 is

Difficult

A

$\frac{1}{14}$
B

$\frac{1}{15}$
C

$\frac{1}{17}$
D

$\frac{1}{16}$

Solution

$P (E) = [\frac{s u m 16}{e x a c t l y 2 n u m b e r s s a m e}]$
= $\frac{6}{{}^{6}C_{1} \times {}^{5}C_{1} \times \frac{3!}{2!}} = \frac{1}{15}$
Sum 16={(6,6,4), (5.5,6)}=3+3=6

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