AMC 10 Daily Practice - Probability

Complete problem set with solutions and individual problem pages

Problem 2 Easy

Two teachers and two students are lined up in a row for a photo. Let A be the number of teachers between the two students. Then E(A)=            .

  • A.

    \frac{1}{4}

  • B.

    \frac{1}{3}

  • C.

    \frac{2}{3}

  • D.

    \frac{4}{3}

  • E.

    \frac{3}{4}

Answer:C

The possible values of the random variable A are 0,1, and 2.

\begin{aligned} & P(A=0)=\frac{2 \times 2+2 \times 2 \times 2}{A_4^4}=\frac{1}{2} \\ & P(A=1)=\frac{C_2^1 \cdot A_2^2 \cdot C_2^1}{A_4^4}=\frac{1}{3} \\ & P(A=2)=\frac{2 \times 2}{A_4^4}=\frac{1}{6} \end{aligned}

Therefore,

E(A)=0 \times \frac{1}{2}+1 \times \frac{1}{3}+2 \times \frac{1}{6}=\frac{2}{3}.

Related Problems
Problem 1 AMC 10 Daily Practice - Probability
Problem 3 AMC 10 Daily Practice - Probability
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