First slide

Random variable and probability distribution

Question

Three dice are rolled 4 times. The probability of getting sum 17 exactly 3 times is

Moderate

A

$\frac{4 \times 71}{72^{4}}$
B

$\frac{71}{72^{4}}$
C

$\frac{70}{72^{4}}$
D

$\frac{1}{72^{4}}$

Solution

total 17 when $(6, 6, 5) (6, 5, 6) (5, 6, 6)$
$p = \frac{3}{6^{3}} = \frac{1}{72}, q = \frac{71}{72}, n = 4$
$P (X = 3) = 4_{C_{3}} \frac{71}{72} . {(\frac{1}{72})}^{^{3}}$
= $\frac{4 \times 71}{72^{4}}$

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