A drawer holds 7 blue socks and 5 red socks. Two socks are taken out at random, one after the other, and the first sock is not put back before the second is taken. What is the probability that both socks are red?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\frac{5}{33}\)
Without replacement, the first red sock stays out, so the second draw is made from one sock fewer and one red sock fewer. The probability of both draws is the product of the two probabilities.
Total socks \(=7+5=12\)
P(first sock red) \(\displaystyle =\frac{5}{12}\)
After that, 11 socks are left, of which 4 are red: P(second sock red) \(\displaystyle =\frac{4}{11}\)
P(both red) \(\displaystyle =\frac{5}{12}\times\frac{4}{11}=\frac{20}{132}=\frac{5}{33}\)