Adam, Bob, and Carol wait in a line, each intending to buy one of the six hats of different colors: red, blue, yellow,

[Vincen]

Adam, Bob, and Carol wait in a line, each intending to buy one of the six hats of different colors: red, blue, yellow, green, black, and white. What is the probability that the blue hat will be bought by Bob or Carol?

A. \(\dfrac1{120}\)

B. \(\dfrac1{24}\)

C. \(\dfrac16\)

D. \(\dfrac15\)

E. \(\dfrac13\)

Answer: E

Source: Economist

[Scott@TargetTestPrep]

Adam, Bob, and Carol wait in a line, each intending to buy one of the six hats of different colors: red, blue, yellow, green, black, and white. What is the probability that the blue hat will be bought by Bob or Carol?

A. \(\dfrac1{120}\)

B. \(\dfrac1{24}\)

C. \(\dfrac16\)

D. \(\dfrac15\)

E. \(\dfrac13\)

Answer: E

Solution:

The probability that Bob buys the blue hat is ⅙, and the probability that Carol buys the blue hat is also 1/6. Thus, the probability that the blue hat will be bought by Bob or Carol is P(Bob or Carol) = P(Bob) + P(Carol) - P(Both Bob and Carol) = 1/6 + 1/6 - 0 = 2/6 = 1/3.

Many test takers make the mistake of thinking one of the probabilities should be 1/5 because if we assume one of Bob or Carol already made the choice, there are now 5 hats remaining for the other person to choose. Notice that this is only true IF we assume the first chosen hat is NOT blue, which has a probability of 5/6. If the first hat is blue, the probability that the second hat is also blue is 0. Thus, the probability that the second hat is blue is actually equal to the product P(first hat is not blue) * P(assuming the first hat is not blue, the second hat is blue). Thus, the probability that the second chosen hat is blue is 5/6 * 1/5 = 1/6. Note that each of the probabilities that the third, fourth, fifth, etc. chosen hat is blue is also 1/6, using a similar reasoning.

Answer: E