A bag contains 6 balls. If three balls are drawn at a time and all of them are found to be green, then the probability that exactly 5 of the balls in the bag are green is:

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Aicharya · Accepted Answer

Let $A$ be the event we draw three green balls. Let $B$ be the event that fine balls are green. Let $E_1$ be event that there are 3 green balls. $E_2$ be event that there are 4 green balls. $E_3$ be event that there are 5 green balls. $E_4$ be event that there are 6 green balls.
$\begin{aligned} & P(A)=P\left(\frac{A}{E_1}\right) \times P\left(E_1\right)+P\left(\frac{A}{E_2}\right) \times P\left(E_2\right) \\ & +P\left(\frac{A}{E_3}\right) \times P\left(E_3\right)+P\left(\frac{A}{E_4}\right) \times P\left(E_4\right) \\ & =\frac{{ }^3 C_3}{{ }^6 C_3} \times\left(\frac{1}{2}\right)^6+\frac{{ }^4 C_3}{{ }^6 C_3} \times\left(\frac{1}{2}\right)^6 \\ & +\frac{{ }^5 C_3}{{ }^6 C_3} \times\left(\frac{1}{2}\right)^6+\frac{{ }^6 C_3}{{ }^6 C_3} \times\left(\frac{1}{2}\right)^6 \end{aligned}$ $\therefore$ Required probability $P\left(\frac{B}{A}\right)=\frac{P\left(\frac{A}{B}\right) \times P(B)}{P(A)}$ $\begin{aligned} & =\frac{\frac{{ }^5 C_3}{{ }^6 C_3} \times\left(\frac{1}{2}\right)^6}{\frac{1}{20}\left(\frac{1}{2}\right)^6+\frac{1}{5}\left(\frac{1}{2}\right)^6+\left(\frac{1}{2}\right)^7+\left(\frac{1}{2}\right)^6} \\ & =\frac{\frac{1}{2}}{\frac{1}{20}+\frac{1}{5}+\frac{1}{2}+1}=\frac{10}{35}=\frac{2}{7} \end{aligned}$

A bag contains 6 balls. If three balls are drawn at a time and all of them are found to be green, then the…

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