Probability Ques 94
- A pot contain 5 red and 2 green balls. At random a ball is drawn from this pot. If a drawn ball is green then put a red ball in the pot and if a drawn ball is red, then put a green ball in the pot, while drawn ball is not replace in the pot. Now we draw another ball randomnly, the probability of second ball to be red is (2019 Main, 9 Jan II)
(a) $\frac{27}{49}$
(b) $\frac{26}{49}$
(c) $\frac{21}{49}$
(d) $\frac{32}{49}$
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Answer:
Correct Answer: 94.(d)
Solution:
- Let $A$ be the event that ball drawn is given and $B$ be the event that ball drawn is red.
$$ \therefore \quad P(A)=\frac{2}{7} \text { and } P(B)=\frac{5}{7} $$
Again, let $C$ be the event that second ball drawn is red.
$$ \begin{aligned} \therefore \quad P(C) & =P(A) \quad P(C / A)+P(B) P \\ & =\frac{2}{7} \times \frac{6}{7}+\frac{5}{7} \times \frac{4}{7} \\ & =\frac{12+40}{49}=\frac{32}{49} \end{aligned} $$
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