Question

Find the probability distribution of the number of successes in two tosses of a die, where a succeess is defined as

number greater than 4

six appears on the die.

Answer 1

When a die is tossed two times, we obtain $(6\times 6)=36$ number of sample points.

Let X be the random variable which denote the number greater than 4 in two tosses of a die. So X may have values 0,1 or 2.

Now, P(X=0)=P (number less than or equal to 4 on both the tosses ) =$\frac{4}{6}\times \frac{4}{6}=\frac{16}{36}=\frac{4}{9},$

P(X=1)=P (number less than or equal to 4 on first toss and greater than 4 on second toss )+ P(number greater than 4 on first toss and less than or equal to 4 on second toss)

$=\frac{4}{6}\times \frac{2}{6}+\frac{4}{6}\times \frac{2}{6}=\frac{16}{36}=\frac{4}{9}$

P(X=2) =P (number greater than 4 on both the tosses) $=\frac{2}{6}\times \frac{2}{6}=\frac{4}{36}=\frac{1}{9}$

Probability distribution of X, i.e., number of successes is

$\begin{array}{cccc}X& 0& 1& 2\\ P\left(X\right)& \frac{4}{9}& \frac{4}{9}& \frac{1}{9}\end{array}$

Let X be the random variable which denotes the number of six appears on atleast one die. So, X may have values 0 or 1.

P(X=0) =P (six does not appear on any of the die)

$=\frac{5}{6}\times \frac{5}{6}=\frac{25}{36}$

P(X=1) =P (six appears on atleast one of the die) $=\frac{11}{36}$

Thus , the required probability distribution is as follows

$\begin{array}{ccc}X& 0& 1\\ P\left(X\right)& \frac{25}{36}& \frac{11}{36}\end{array}$