An urn contains 5 red and 5 black balls. A ball is drawn at random, its color is noted and is returned to the urn. Moreover, 2 additional balls of the same color are put in the urn and then a ball is drawn at random. The probability that the second drawn ball is red, is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{1}{2}\)
Let \(E_1\) and \(E_2\) be the events that the first ball is red and black, and A the event that the second ball is red.
\( \displaystyle P(E_1) = P(E_2) = \frac{5}{10} = \frac{1}{2} \)
If the first ball is red, the urn then has 7 red and 5 black balls; if black, 5 red and 7 black:
\( \displaystyle P(A|E_1) = \frac{7}{12},\quad P(A|E_2) = \frac{5}{12} \)
By the theorem of total probability:
\( \displaystyle P(A) = \frac{1}{2}\cdot\frac{7}{12} + \frac{1}{2}\cdot\frac{5}{12} \)
\( \displaystyle = \frac{12}{24} = \frac{1}{2} \)
Tip: When the drawn ball is returned together with extra balls of the same colour, the probability that the next ball is red stays equal to the original proportion of red balls, here \(\frac{5}{10} = \frac{1}{2}\).
NCERT (2026-27 edition):
Bayes Theorem
