Suppose we have four boxes A, B, C and D containing coloured marb… - Vidyadip

Question

Suppose we have four boxes A, B, C and D containing coloured marbles as given below:

Box

Marble colour

Red

White

Black

A

B

C

D

1

6

8

0

6

2

1

6

3

2

1

4

One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A?, box B? box C?

✓

Answer

Let R represents the drawing of red ball and the four boxes are represented by A, B, C and D.
So $P(R|A) = \frac{1}{{10}}$, $P(R|B) = \frac{6}{{10}}$,
$P(R|C) = \frac{8}{{10}},P(R|D) = \frac{0}{{10}} = 0$
Since there are 4 bags.
Therefore, P (A) = $\frac{1}{4},$ P (B) = $\frac{1}{4},$ P (C) = $\frac{1}{4},$ P (D) = $\frac{1}{4}$
$P(A|R) = \frac{{P(A) \cdot P(R|A)}}{{P(A) \cdot P(R|A) + P(B) \cdot P(R|B) + P(C) \cdot P(R|C) + P(D) \cdot (R|D)}}$
$ = \frac{{\frac{1}{4} \times \frac{1}{{10}}}}{{\frac{1}{4} \times \frac{1}{{10}} + \frac{1}{4} \times \frac{6}{{10}} + \frac{1}{4} \times \frac{8}{{10}} + \frac{1}{4} \times 0}}$
$ = \frac{{\frac{1}{{10}}}}{{\frac{1}{{10}} + \frac{6}{{10}} + \frac{8}{{15}}}} = \frac{1}{{15}}$
$P(B/R) = \frac{{P(B) \cdot P(R|B)}}{{P(A)P(R|A) + P(B)P(R|B) + P(C)P(R|C) + P(D)P(R|D)}}$
$ = \frac{{\frac{1}{4} \times \frac{6}{{10}}}}{{\frac{1}{4} \times \frac{1}{{10}} + \frac{1}{4} \times \frac{6}{{10}} + \frac{1}{4} \times \frac{8}{{10}} + \frac{1}{4} \times 0}} = \frac{6}{{15}} = \frac{2}{5}$
$P(C|R) = \frac{{P(C) \cdot P(R|C)}}{{P(A)P(R|A) + P(B)P(R|B) + P(C)P(R|C) + P(D)P(R|D)}}$
$ = \frac{{\frac{1}{4} \times \frac{8}{{10}}}}{{\frac{1}{4} \times \frac{1}{{10}} + \frac{1}{4} \times \frac{6}{{10}} + \frac{1}{4} \times \frac{8}{{10}} + \frac{1}{4} \times 0}} = \frac{8}{{15}}$

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