A bag contains 6 balls. If three balls are dr | vedclass

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Question

(C) Let $A$ be the event that we draw three green balls. Let $E_k$ be the event that there are $k$ green balls in the bag,where $k \in \{3, 4, 5, 6\}$

Vedclass · Accepted Answer

$\frac{2}{7}$

Vedclass · Answer

(C) Let $A$ be the event that we draw three green balls. Let $E_k$ be the event that there are $k$ green balls in the bag,where $k \in \{3, 4, 5, 6\}$. Assuming each number of green balls is equally likely,$P(E_k) = \frac{1}{4}$.The probability of drawing $3$ green balls given $k$ green balls are present is $P(A|E_k) = \frac{{}^k C_3}{{}^6 C_3}$.Using Bayes' Theorem,the probability that there are $5$ green balls given that $3$ green balls were drawn is:$P(E_5|A) = \frac{P(A|E_5)P(E_5)}{\sum_{k=3}^{6} P(A|E_k)P(E_k)}$Since $P(E_k) = \frac{1}{4}$ for all $k$,this simplifies to:$P(E_5|A) = \frac{{}^5 C_3}{\sum_{k=3}^{6} {}^k C_3} = \frac{10}{1 + 4 + 10 + 20} = \frac{10}{35} = \frac{2}{7}$.

$A$ 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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