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Q1:
An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?

Solution :

Step 1: Define Events and Initial Probabilities

Let $E_1$ be the event that the first ball drawn is Red, and $E_2$ be the event that the first ball drawn is Black. The urn initially contains 5 Red ($R$) and 5 Black ($B$) balls, totaling 10 balls.

  • $P(E_1) = \frac{5}{10} = \frac{1}{2}$
  • $P(E_2) = \frac{5}{10} = \frac{1}{2}$
[By the definition of classical probability: $P(Event) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$]

Step 2: Define Conditional Probabilities for the Second Draw

Let $A$ be the event that the second ball drawn is Red. We must consider the state of the urn after the first ball is replaced and two additional balls of the same color are added.

Scenario Urn Composition (after replacement + 2 added) Probability of Second Ball being Red ($P(A|E_i)$)
First ball is Red ($E_1$) $5+2=7$ Red, $5$ Black (Total = 12) $P(A|E_1) = \frac{7}{12}$
First ball is Black ($E_2$) $5$ Red, $5+2=7$ Black (Total = 12) $P(A|E_2) = \frac{5}{12}$

Step 3: Apply the Law of Total Probability

To find the total probability of event $A$, we use the Law of Total Probability: $P(A) = P(E_1) \cdot P(A|E_1) + P(E_2) \cdot P(A|E_2)$

Substituting the values derived above:

$P(A) = \left( \frac{1}{2} \times \frac{7}{12} \right) + \left( \frac{1}{2} \times \frac{5}{12} \right)$

Step 4: Algebraic Simplification

$P(A) = \frac{7}{24} + \frac{5}{24}$

$P(A) = \frac{12}{24}$

$P(A) = \frac{1}{2}$

Red: 7/12 Red: 5/12 Start Red (1/2) Black (1/2)

Final Solution: The probability that the second ball is red is $\frac{1}{2}$.


More Questions from Class 12 Mathematics Probability EXERCISE 13.3


CBSE Solutions for Class 12 Mathematics Probability


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