Question

An urn contains $5$ red and $2$ green balls. A ball is drawn at random from the urn. If the drawn ball is green, then a red ball is added to the urn and if the drawn ball is red, then a green ball is added to the urn; the original ball is not returned to the urn. Now, a second ball is drawn at random from it. The probability that the second ball is red, is :

• A. $\frac{26}{49}$
• B. $\frac{32}{49}$
• C. $\frac{27}{49}$
• D. $\frac{21}{49}$

Answer 1

The Correct Option is B

To solve this problem, let's break it down into parts and calculate the probabilities step by step.

Initially, the urn contains 5 red balls and 2 green balls, summing up to a total of 7 balls. Therefore, the probability of drawing a red ball first is:

\(\frac{5}{7}\)

And the probability of drawing a green ball first is:

\(\frac{2}{7}\)

Case 1: The first ball drawn is red.

• If the drawn ball is red, then a green ball is added to the urn. The composition of the urn becomes 4 red balls and 3 green balls.
• Now, the probability of drawing a red ball in the second draw from the urn (which now contains 7 balls again) is:
• \(\frac{4}{7}\)

Case 2: The first ball drawn is green.

• If the drawn ball is green, then a red ball is added to the urn. The composition of the urn becomes 6 red balls and 1 green ball.
• Now, the probability of drawing a red ball in the second draw from the urn (which again contains 7 balls) is:
• \(\frac{6}{7}\)

Now, we need to find the total probability of drawing a red ball in the second draw by considering both cases:

• Probability of drawing a red ball first and then another red ball: \(\frac{5}{7} \times \frac{4}{7} = \frac{20}{49}\)
• Probability of drawing a green ball first and then a red ball: \(\frac{2}{7} \times \frac{6}{7} = \frac{12}{49}\)

Thus, the total probability of drawing a red ball on the second draw is:

\(\frac{20}{49} + \frac{12}{49} = \frac{32}{49}\)

Therefore, the probability that the second ball is red is \(\frac{32}{49}\).