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Probability

Section: Mathematics

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Question : 8 of 92

Marks: +1, -0

A bag contains 5 balls of unknown colours. There are equal chances that out of these five balls, there may be 0 or 1 or 2 or 3 or 4 or 5 red balls. A ball is taken out from the bag at random and is found to be red. The probability that it is the only red ball in the bag is

[AP EAPCET 23 May 2025 S2]

$\frac{1}{5}$
$\frac{1}{6}$
$\frac{1}{15}$
$\frac{1}{30}$

Solution:

Let

R_k

be the event that are exactly

k

red balls is in bag, where

k=1,2,3,4,5

So,

P(R_k) = \frac{1}{5}

for

k=1,2,3,4,5

Let

A

be the event that a red ball is drawn.

P(R_1 / A) = \frac{P(R_1 \cap A)}{P(A)}

\because\ P(A / R_1) = \frac{1}{5}

\text{ and }\ P(R_1 \cap A) = P(R_1) \cdot P(A / R_1)

= \frac{1}{5} \cdot \frac{1}{5} = \frac{1}{25}

\text{ and }\ P(A) = \sum\limits_{k=0}^{5} P(R_k) \cdot P(A / R_k)

= \frac{1}{5} \sum\limits_{k=0}^{5} \frac{k}{5} = \frac{1}{25}(1+2+3+4+5)

= \frac{1}{25} \times 15 = \frac{15}{25}

Hence,

P(R_1 / A) = \frac{P(R_1 \cap A)}{P(A)} = \frac{\frac{1}{25}}{\frac{15}{25}} = \frac{1}{15}

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