Question

find the probability distribution of the number of sixes in throwing three dice once.

• A. $\frac{1}{216}$
• B. $\frac{1}{6}$
• C. $\frac{1}{16}$
• D. $\frac{1}{26}$

Answer 1

The correct option is A $\frac{1}{216}$

Let $X$ be the discrete denoting the number of sixes when three dice are thrown once

$X$ takes value $0,1,2,3$

Step $2:-$

$P(X=0)=$ Probability that all three dice do not show $6=\frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}=\frac{125}{216}$

$P(X=1)=$ Probability of one dices having $6$

$=3C_1\left(\frac{1}{6}\right)\times \left(\frac{5}{6}\right)\times \left(\frac{5}{6}\right)=\frac{75}{216}=\frac{25}{72}$

$P(X=2)=$ Probability of two dice showing $6$

$=3C_2\left(\frac{1}{6}\right)\times \left(\frac{1}{6}\right)\times \left(\frac{5}{6}\right)=\frac{15}{216}=\frac{5}{72}$

$P(X=3)=$ Probability of these dice showing $6$

$=\frac{1}{6}\times \frac{1}{6}\times \frac{1}{6}=\frac{1}{216}$

Step $3$

The probability distribution is given by

$(X=X)0123P(X=X)\frac{125}{216}\frac{25}{72}\frac{5}{72}\frac{1}{216}$