Conditional Probability: Application Problems (1)

easy • 2 min read
Tags Conditional Probability

Question

A bag contains 5 red and 3 blue balls. Two balls are drawn one after the other without replacement. (a) Find the probability that both balls are red. (b) Given that the first ball drawn is red, find the probability that the second is also red. (c) Find the probability that the second ball is red.

Solution — Step by Step

P(first red)=5/8P(\text{first red}) = 5/8. After drawing one red, 44 red and 33 blue remain. So P(second red∣first red)=4/7P(\text{second red} \mid \text{first red}) = 4/7. By multiplication rule:

P(both red)=58×47=2056=514P(\text{both red}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}

Already used above: P=4/7P = 4/7.

Use the law of total probability:

P(second red)=P(1st R)P(2nd R∣1st R)+P(1st B)P(2nd R∣1st B)P(\text{second red}) = P(\text{1st R})P(\text{2nd R}\mid\text{1st R}) + P(\text{1st B})P(\text{2nd R}\mid\text{1st B}) =58⋅47+38⋅57=2056+1556=3556=58= \frac{5}{8}\cdot\frac{4}{7} + \frac{3}{8}\cdot\frac{5}{7} = \frac{20}{56} + \frac{15}{56} = \frac{35}{56} = \frac{5}{8}

P(both red)=5/14P(\text{both red}) = 5/14, P(2nd R∣1st R)=4/7P(\text{2nd R}\mid\text{1st R}) = 4/7, P(2nd R)=5/8P(\text{2nd R}) = 5/8.

Why This Works

Conditional probability P(A∣B)=P(A∩B)/P(B)P(A\mid B) = P(A \cap B)/P(B) updates our probabilities given partial information. Drawing without replacement creates exactly this dependency: the result of the first draw changes the composition of the bag.

The unconditional second-draw probability equals the first-draw probability (5/85/8 each) — a beautiful symmetry that does not require us to know which one came first.

Alternative Method

For part (c), think of it this way: by symmetry, each ball in the bag is equally likely to occupy the “second” slot. So P(second red)=5/8P(\text{second red}) = 5/8, the same as the original proportion. This is the “exchangeability” argument and saves time on JEE Main MCQs.

Common Mistake

Treating the two draws as independent. They are not — without replacement, the first draw changes the conditional distribution of the second. Students sometimes write P(both red)=(5/8)2P(\text{both red}) = (5/8)^2, which is wrong by exactly the factor that distinguishes “with” from “without” replacement.

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