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Bayes' Theorem

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 II.
One of the two boxes, box I and box II was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability that this red ball was drawn from box II, is $\frac{1}{3},$ then the correct option(s) with the possible values of $n_{1}, n_{2}, n_{3}$ and $n_{4}$ is/are

A

n_{1} = 3, n_{2} = 3, n_{3} = 5, n_{4} = 15

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B

n_{1} = 3, n_{2} = 6, n_{3} = 10, n_{4} = 50

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C

n_{1} = 8, n_{2} = 6, n_{3} = 5, n_{4} = 20

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D

n_{1} = 6, n_{2} = 12, n_{3} = 5, n_{4} = 20

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Solution

The correct options are
A $n_{1} = 3, n_{2} = 3, n_{3} = 5, n_{4} = 15$
B $n_{1} = 3, n_{2} = 6, n_{3} = 10, n_{4} = 50$

Let A = Drawing red ball
$∴ P (A) = P (B_{1}) . P (A | B_{1}) + P (B_{2}) . P (A | B_{2}) = \frac{1}{2} (\frac{n_{1}}{n_{1} + n_{2}}) + \frac{1}{2} (\frac{n_{3}}{n_{3} + n_{4}})$
Given, $P (B_{2} | A) = \frac{1}{3}$
$\Rightarrow \frac{P (B_{2} \cap A)}{P (A)} = \frac{1}{3} \Rightarrow \frac{P (B_{2}) . P (A | B_{2})}{P (A)} = \frac{1}{3} \Rightarrow \frac{\frac{1}{2} (\frac{n_{3}}{n_{3} + n_{4}})}{\frac{1}{2} (\frac{n_{1}}{n_{1} + n_{2}}) + \frac{1}{2} (\frac{n_{3}}{n_{3} + n_{4}})} = \frac{1}{3}$
options A and B satisfy this condition.

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Similar questions

n_{1}

n_{2}

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

n_{3}

n_{4}

be the number of red and black balls, respectively in box II.
One of the two boxes, box I and box II was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability that this red ball was drawn from box II, is

\frac{1}{3},

then the correct option(s) with the possible values of

n_{1}, n_{2}, n_{3}

n_{4}

n_{1}

n_{2}

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

n_{3}

n_{4}

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

\frac{1}{3}

, then the correct option(s) with the possible values of

n_{1}

n_{2}

n_{1}

n_{2}

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

I

n_{3}

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}

n_{2}

Q. One of the two boxes, box I and box II, was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability that this red ball was drawn from box II is

\frac{1}{3}

, then the correct option(s) with the possible values of

n_{1}, n_{2}, n_{3}

n_{4}

n_{1}

n_{2}

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

n_{3}

n_{4}

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

\frac{1}{3}

, then the correct option(s) with the possible values of

n_{1}

n_{2}

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