Two dice are thrown, and $X$ is the larger of the two numbers (either, if they are equal). $P(X = k) = \frac{2k - 1}{36}$ . What is $E(X)$ ?

\frac{91}{36}

Correct answerYour answer

\frac{125}{36}

Correct answerYour answer

\frac{7}{2}

Correct answerYour answer

\frac{161}{36}

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer D

Correct answer: D

$E(X) = \frac{1 \times 1 + 2 \times 3 + 3 \times 5 + 4 \times 7 + 5 \times 9 + 6 \times 11}{36}$ .
$= \frac{161}{36} \approx 4.47$ .

Worth noting

3.5 is the mean of one die; the larger of two is likely to be higher.

Ten questions from Random Variables, Expectation & Variance, at your own pace.

Ten questions from Random Variables, Expectation & Variance, at your own pace.

Two dice are thrown, and XXX is the larger of the two numbers (either, if they are equal). P(X=k)=2k−136P(X = k) = \frac{2k - 1}{36}P(X=k)=362k−1​. What is E(X)E(X)E(X)?