91Ó°ÊÓ

It is given that \({p_1} = 0.11,\,\,{p_2} = 0.30,\,\,{p_3} = 0.22,\,\,{p_4} = 0.05,\,\,{p_5} = 0.25\,\,and\,{p_6} = 0.07.\)These are the probability values of different numbers that appear on a die, that is , 1,2,3,4,5 and 6.

The die is rolled 40 times

Here two events are described:

\({X_1}\): Denotes the number of rolls for which an even number appears

\({X_2}\): Denotes the event when either number 1 or 3 occurs.

The probability that on a roll an even number occurs is

\(\begin{array}{l} = {p_2} + {p_4} + {p_6}\\ = 0.3 + 0.05 + 0.07\\ = 0.42\end{array}\)

Therefore, the distribution is

\({X_1} \sim Bin\left( {n = 40,p = 0.42} \right)\)

The probability that on a roll an either 1 or 3 occurs is

\(\begin{array}{l} = {p_1} + {p_3}\\ = 0.11 + 0.22\\ = 0.33\end{array}\)

Therefore, the distribution is

\({X_2} \sim Bin\left( {n = 40,p = 0.33} \right)\)

We need to find:

\(\begin{array}{l}\Pr \left( {{X_1} = 20\,\,and\,\,{X_2} = 15} \right)\\ = {}^{40}{C_{20}}{\left( {0.42} \right)^{20}}{\left( {0.58} \right)^{20}} \times {}^{40}{C_{15}}{\left( {0.33} \right)^{15}}{\left( {0.67} \right)^{25}}\\ = 8.0727 \times {10^{ - 3}}\end{array}\)

Hence the required solution is\({\bf{8}}{\bf{.0727 \times 1}}{{\bf{0}}^{{\bf{ - 3}}}}\)