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Simple and Compound Event

Let n 1 and n...

Question

Let $n_{1}$ and $n_{2}$ be the number of red and black balls, respectively, in box $I$ . Let $n_{3}$ and $n_{4}$ be the number of red and black balls, respectively, in box $I I$ .

A ball is drawn at random from box $I$ and transferred to box $I I$ . If the probability of drawing a red ball from box $I$ , after this transfer, is $\frac{1}{3}$ , then the correct option(s) with the possible values of $n_{1}$ and $n_{2}$ is(are)

A

n_{1} = 4

and

n_{2} = 6

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B

n_{1} = 2

and

n_{2} = 3

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C

n_{1} = 10

and

n_{2} = 20

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D

n_{1} = 3

and

n_{2} = 6

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Solution

The correct option is D $n_{1} = 3$ and $n_{2} = 6$
Box $I :$
Red balls $- n_{1}$ and black balls $- n_{2}$
$P (R) = \frac{n_{1}}{n_{1} + n_{2}} \cdot \frac{n_{1} - 1}{n_{1} + n_{2} - 1} + \frac{n_{2}}{n_{1} + n_{2}} \cdot \frac{n_{1}}{n_{1} + n_{2} - 1} \Rightarrow \frac{1}{3} = \frac{n_{1}^{2} - n_{1} + n_{1} n_{2}}{(n_{1} + n_{2})^{2} - (n_{1} + n_{2})} \Rightarrow (n_{1} + n_{2}) (n_{1} + n_{2} - 1) = 3 n_{1} (n_{1} + n_{2} - 1) \Rightarrow 2 n_{1} = n_{2}$

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be the number of red and black balls, respectively in box I. Let

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A ball is drawn at random from box I and transferred to box II. If the probability of drawing a red ball from box I, after this transfer, is

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