In a workshop, there are five machines and the probability of any one of them to be out of service on a day is ${{1 \over 4}}$ . If the probability that at most two machines will be out of service on the same day is ${\left( {{3 \over 4}} \right)^3}k$, then k is equal to :
Solution
Probablity of at most two machines will be out of service = ${\left( {{3 \over 4}} \right)^3}k$
<br><br>$\Rightarrow$ <sup>5</sup>C<sub>0</sub>${\left( {{3 \over 4}} \right)^5}$ + <sup>5</sup>C<sub>1</sub>$\left( {{1 \over 4}} \right){\left( {{3 \over 4}} \right)^5}$ + <sup>5</sup>C<sub>2</sub>${\left( {{1 \over 4}} \right)^2}{\left( {{3 \over 4}} \right)^5}$ = ${\left( {{3 \over 4}} \right)^3}k$
<br><br>$\Rightarrow$ ${{17} \over 8}{\left( {{3 \over 4}} \right)^3}$ = ${\left( {{3 \over 4}} \right)^3}k$
<br><br>$\Rightarrow$ k = ${{17} \over 8}$
About this question
Subject: Mathematics · Chapter: Probability · Topic: Classical and Axiomatic Probability
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