Question

Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as
(i) number greater than 4
(ii) six appears on at least one die

Answer 1

When a die is tossed two times, we obtain $(6\times 6)=36$ number of observations.
Let $X$ be the random variable, which represents the number of successes.
(i) Here, success refers to the number greater than 4.
$P(X=0)=P\left(\text{number less than or equal to 4 on both the tosses}\right)=\frac{4}{6}\times \frac{4}{6}=\frac{4}{9}$
$P(X=1)=P\left(\text{number less than or equal to 4 on first toss and greater than 4 on second toss}\right)+$

$P\left(\text{number greater than 4 on first toss and less than or equal to 4 on second toss}\right)$

$=\frac{4}{6}\times \frac{2}{6}+\frac{4}{6}\times \frac{2}{6}=\frac{4}{9}$
$P(X=2)=P\left(\text{number greater than 4 on both the tosses}\right)=\frac{2}{6}\times \frac{2}{6}=\frac{1}{9}$
Thus, the probability distribution is as follows.
(ii) Here, success means six appears on at least one die.
$P(Y=0)=P\left(\text{six does not appear on any of the dice}\right)=\frac{5}{6}\times \frac{5}{6}=\frac{25}{36}$
$P(Y=1)=P\left(\text{six appears on at least one of the dice}\right)=\frac{11}{36}$
Thus, the required probability distribution is as follows.