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(D) When two dice are thrown, the total number of outcomes is $6 	imes 6 = 36$. The sum of the numbers on the two dice can be even or odd. The number

Vedclass · Accepted Answer

$\frac{3}{4}$

Vedclass · Answer

(D) When two dice are thrown, the total number of outcomes is $6 	imes 6 = 36$. The sum of the numbers on the two dice can be even or odd. The number of outcomes where the sum is even is $18$, and the number of outcomes where the sum is odd is $18$. Thus, $P(	ext{sum is even}) = \frac{18}{36} = \frac{1}{2}$ and $P(	ext{sum is odd}) = \frac{18}{36} = \frac{1}{2}$. Let $X$ be the random variable representing the points obtained by $A$. If the sum is even, $X = \frac{1}{2}$ with probability $\frac{1}{2}$. If the sum is odd, $X = 1$ with probability $\frac{1}{2}$. The arithmetic mean (expected value) of $X$ is $E(X) = \sum x_i p_i = (\frac{1}{2} 	imes \frac{1}{2}) + (1 	imes \frac{1}{2}) = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}$.

$X$	$1$	$2$	$3$	$4$
$P(X)$	$\frac{1}{10}$	$\frac{1}{5}$	$\frac{3}{10}$	$\frac{2}{5}$

$X$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X)$	$0$	$k$	$2k$	$3k$	$3k^2$	$k^2$	$2k^2$	$7k^2+k$

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