2019 AMC 12A Problem 20

Question

Real numbers between

0

and

1,

inclusive, are chosen in the following manner. A fair coin is flipped. If it lands heads, then it is flipped again and the chosen number is

0

if the second flip is heads and

1

if the second flip is tails. On the other hand, if the first coin flip is tails, then the number is chosen uniformly at random from the closed interval

[0, 1].

Two random numbers

x

and

y

are chosen independently in this manner. What is the probability that

|x - y| \gt \dfrac{1}{2}?

\dfrac{1}{3}

\dfrac{7}{16}

\dfrac{1}{2}

\dfrac{9}{16}

\dfrac{2}{3}

Answer: B

Concepts:geometric probability casework

Difficulty rating: 2070

Small hint:

Each number is

0

with probability

\tfrac14,

1

with probability

\tfrac14,

and uniform on

[0,1]

with probability

\tfrac12

Big hint:

Split into the nine type-combinations; for two uniforms,

P\!\left(|x - y| \gt \tfrac12\right) = \tfrac14

Solution:

Each variable equals

0

with probability

\tfrac14,

equals

1

with probability

\tfrac14,

and is uniform on

[0, 1]

with probability

\tfrac12.

Considering the nine combinations of types: the pairs

(0, 1)

and

(1, 0)

each contribute

\tfrac{1}{16}.

Each of the four point-versus-uniform cases contributes

\tfrac{1}{16}.

The uniform-versus-uniform case contributes

\tfrac14 \cdot \tfrac14 = \tfrac{1}{16}.

The total is

\dfrac{2 + 4 + 1}{16} = \dfrac{7}{16}.

Thus, the correct answer is B.

2019 AMC 12A Problem 20

Answer: B

Difficulty rating: 2070

Small hint:

Big hint:

Solution:

Problem 20 in other years