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Suppose that a die is loaded so that each of the numbers 1, 2, 3, 4, 5, and 6 has a different probability of appearing when the die is rolled. For\({\bf{i = 1, \ldots ,6,}}\)let\({{\bf{p}}_{\bf{i}}}\)denote the probability that the number i will be obtained, and

suppose that\({{\bf{p}}_{\bf{1}}}{\bf{ = 0}}{\bf{.11,}}\,\,{{\bf{p}}_{\bf{2}}}{\bf{ = 0}}{\bf{.30,}}\,\,{{\bf{p}}_{\bf{3}}}{\bf{ = 0}}{\bf{.22,}}\,\,{{\bf{p}}_{\bf{4}}}{\bf{ = 0}}{\bf{.05,}}\,\,{{\bf{p}}_{\bf{5}}}{\bf{ = 0}}{\bf{.25}}\,\,{\bf{and}}\,{{\bf{p}}_{\bf{6}}}{\bf{ = 0}}{\bf{.07}}{\bf{.}}\)Suppose also that the die is to be rolled 40 times. Let\({{\bf{X}}_{\bf{1}}}\)denote the number of rolls for which an even number appears, and let\({{\bf{X}}_2}\)denote the number of rolls for which either the number 1 or the number 3 appears. Find the value of\({\bf{Pr}}\left( {{{\bf{X}}_{\bf{1}}}{\bf{ = 20}}\,\,{\bf{and}}\,\,{{\bf{X}}_{\bf{2}}}{\bf{ = 15}}} \right)\).

Short Answer

Expert verified

\(8.0727 \times {10^{ - 3}}\)

Step by step solution

01

Given information

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.

02

Define the pdf

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

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

03

Calculate the desired probability

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\({\bf{8}}{\bf{.0727 \times 1}}{{\bf{0}}^{{\bf{ - 3}}}}\)

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Most popular questions from this chapter

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