Probability: Tricky Questions Solved (1)

Question

A bag contains 4 red balls and 6 blue balls. Two balls are drawn at random without replacement. Find the probability that both are red.

Solution — Step by Step

Why This Works

When drawing without replacement, each outcome is equally likely if we’re using combinations to count. We could also use the multiplication rule: probability that the first ball is red is $4/10$; given that, probability that the second is also red is $3/9$. Multiplying: $(4/10)(3/9) = 12/90 = 2/15$. Same answer, different route.

The combinations approach treats the order as irrelevant. The sequential approach respects order but cancels the ordering when we ask about events that don’t depend on order.

Alternative Method

Sequential probability:

$P(\text{both red}) = P(\text{first red}) \times P(\text{second red} | \text{first red}) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}$

This is faster when we have just two draws and is the recommended approach for conditional-style problems.

Common Mistake

Treating “without replacement” as “with replacement.” If the problem said with replacement, the second draw resets the bag, so $P = (4/10)(4/10) = 16/100 = 4/25$. Different answer. Always check the wording.