2002 AMC 12A Problem 16

Question

Tina randomly selects two distinct numbers from the set

\{1, 2, 3, 4, 5\},

and Sergio randomly selects a number from the set

\{1, 2, \ldots, 10\}.

The probability that Sergio’s number is larger than the sum of the two numbers chosen by Tina is

\dfrac{2}{5}

\dfrac{9}{20}

\dfrac{1}{2}

\dfrac{11}{20}

\dfrac{24}{25}

Tina has

\binom{5}{2} = 10

equally likely pairs

For a pair with sum

s,

Sergio exceeds it with probability

\dfrac{10 - s}{10}

Tina’s ten pairs have sums

3, 4, 5, 5, 6, 6, 7, 7, 8, 9.

For a sum

s,

Sergio’s number exceeds it with probability

\dfrac{10 - s}{10}.

The corresponding values of

10 - s

are

7, 6, 5, 5, 4, 4, 3, 3, 2, 1,

totaling

40.

The overall probability is

\dfrac{40}{10\cdot 10} = \dfrac{2}{5}.

Thus, the correct answer is A.

Problem 16 in other years