CUET Mathematics 2024 - There are two bags. Bag-1 contains 4 white and 6 black balls and Bag-2 contains 5 white and 5 black balls. A die is rolled, if it shows a number divisible by 3, a ball is drawn from Bag-1, else a ball is drawn from Bag-2. If the ball drawn is not black in colour, the probability that it was not drawn from Bag-2 is : | PYQs + Solutions

CUET Mathematics 2024

Algebra

Probability

Medium

There are two bags. Bag-1 contains $4$ white and $6$ black balls and Bag-2 contains $5$ white and $5$ black balls. A die is rolled, if it shows a number divisible by 3, a ball is drawn from Bag-1, else a ball is drawn from Bag-2. If the ball drawn is not black in colour, the probability that it was not drawn from Bag-2 is :

\frac{4}{9}

\frac{3}{8}

\frac{2}{7}

\frac{4}{19}

✅ Correct Option: 3

Define our events:

\newline

- Event A: Ball is drawn from Bag-1

\newline

- Event B: Ball is white (not black)

We need to find

P(A|B)

- the probability that the ball came from Bag-1 given that it's white.

Using Bayes' Theorem:

P(A|B) = \dfrac{P(B|A) \times P(A)}{P(B)}

Finding each component:

P(A)

= Probability of drawing from Bag-1 = Probability of rolling a number divisible by 3

Numbers divisible by 3 on a die: 3, 6 (2 numbers out of 6)

P(A) = \dfrac{2}{6} = \dfrac{1}{3}

P(B|A)

= Probability of drawing a white ball given we're drawing from Bag-1

Bag-1 has 4 white balls out of 10 total balls

P(B|A) = \dfrac{4}{10} = \dfrac{2}{5}

P(B)

= Total probability of drawing a white ball

P(B) = P(B|A) \times P(A) + P(B|A^c) \times P(A^c)

Where

P(A^c) = 1 - \dfrac{1}{3} = \dfrac{2}{3}

(probability of drawing from Bag-2)

P(B|A^c) = \dfrac{5}{10} = \dfrac{1}{2}

(probability of white ball from Bag-2)

P(B) = \dfrac{2}{5} \times \dfrac{1}{3} + \dfrac{1}{2} \times \dfrac{2}{3} = \dfrac{2}{15} + \dfrac{1}{3} = \dfrac{2}{15} + \dfrac{5}{15} = \dfrac{7}{15}

Apply Bayes' Theorem:

P(A|B) = \dfrac{P(B|A) \times P(A)}{P(B)} = \dfrac{\dfrac{2}{5} \times \dfrac{1}{3}}{\dfrac{7}{15}} = \dfrac{\dfrac{2}{15}}{\dfrac{7}{15}} = \dfrac{2}{7}

Therefore, the probability that the white ball was drawn from Bag-1 is

\dfrac{2}{7}

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