NIMCET 2023 — Mathematics PYQ

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Question

NIMCET 2023 — Mathematics PYQ

NIMCET | Mathematics | 2023

Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. One balls is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be black in colour. Then the probability, that the transferred is red, is:

Choose the correct answer:

Explanation

Step 1: Define the Events

Let the events of transferring a ball from Bag I to Bag II be:

$E_{1}$ : The transferred ball is Red.
$E_{2}$ : The transferred ball is Black.
$E_{3}$ : The transferred ball is White.

Let $A$ be the event that the ball drawn from Bag II is Black.

Step 2: Find the Initial Probabilities of Transfer (Bag I)

Bag I has a total of $3 + 4 + 3 = 10$ balls.

Probability of transferring a Red ball:
$P (E_{1}) = \frac{3}{10}$
Probability of transferring a Black ball:
$P (E_{2}) = \frac{4}{10}$
Probability of transferring a White ball:
$P (E_{3}) = \frac{3}{10}$

Step 3: Find the Conditional Probabilities (Bag II)

Bag II originally has 2 red, 5 black, and 2 white balls (Total = 9 balls). After transferring 1 ball from Bag I, the total number of balls in Bag II becomes $9 + 1 = 10$ .

Case 1: If a Red ball is transferred ( $E_{1}$ )
Bag II now has 3 red, 5 black, and 2 white balls.
$P (A ∣ E_{1}) = \frac{5}{10}$
Case 2: If a Black ball is transferred ( $E_{2}$ )
Bag II now has 2 red, 6 black, and 2 white balls.
$P (A ∣ E_{2}) = \frac{6}{10}$
Case 3: If a White ball is transferred ( $E_{3}$ )
Bag II now has 2 red, 5 black, and 3 white balls.
$P (A ∣ E_{3}) = \frac{5}{10}$

Step 4: Apply Bayes' Theorem

We need to find the probability that the transferred ball was red, given that the drawn ball from Bag II is black. This is represented as $P (E_{1} ∣ A)$ .

According to Bayes' Theorem:

$P (E_{1} ∣ A) = \frac{P ( E _{1} ) \cdot P ( A ∣ E _{1} )}{P ( E _{1} ) \cdot P ( A ∣ E _{1} ) + P ( E _{2} ) \cdot P ( A ∣ E _{2} ) + P ( E _{3} ) \cdot P ( A ∣ E _{3} )}$

Substitute the values into the formula:

$P (E_{1} ∣ A) = \frac{\frac{3}{10} \times \frac{5}{10}}{( \frac{3}{10} \times \frac{5}{10} ) + ( \frac{4}{10} \times \frac{6}{10} ) + ( \frac{3}{10} \times \frac{5}{10} )}$

Simplify the expression by canceling out the common denominator $100$ :

$P (E_{1} ∣ A) = \frac{3 \times 5}{( 3 \times 5 ) + ( 4 \times 6 ) + ( 3 \times 5 )}$

$P (E_{1} ∣ A) = \frac{15}{15 + 24 + 15}$

$P (E_{1} ∣ A) = \frac{15}{54}$

Reducing the fraction to its simplest form by dividing the numerator and the denominator by 3:

$P (E_{1} ∣ A) = \frac{5}{18}$

Conclusion

The correct option is B) $\frac{5}{18}$ .