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(B) We know that the sum of probabilities for all possible values of a random variable is $1$. $\sum_{k=0}^{5} P(X=k) = 1$ $a \left( \frac{0+1}{2^0} +

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$\frac{23}{60}$

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(B) We know that the sum of probabilities for all possible values of a random variable is $1$. $\sum_{k=0}^{5} P(X=k) = 1$ $a \left( \frac{0+1}{2^0} + \frac{1+1}{2^1} + \frac{2+1}{2^2} + \frac{3+1}{2^3} + \frac{4+1}{2^4} + \frac{5+1}{2^5} ight) = 1$ $a \left( 1 + 1 + \frac{3}{4} + \frac{4}{8} + \frac{5}{16} + \frac{6}{32} ight) = 1$ $a \left( 2 + 0.75 + 0.5 + 0.3125 + 0.1875 ight) = 1$ $a \left( \frac{64+48+32+20+12}{32} ight) = 1$ $\Rightarrow a \left( \frac{176}{32} ight) = 1$ $\Rightarrow a \left( \frac{11}{2} ight) = 1$ $\Rightarrow a = \frac{2}{11}$. Wait,re-calculating the sum: $a(1 + 1 + 0.75 + 0.5 + 0.3125 + 0.1875) = a(2 + 0.75 + 0.5 + 0.5) = a(3.75) = a(\frac{15}{4}) = 1$ $\Rightarrow a = \frac{4}{15}$. The prime values for $X$ in the set $\{0, 1, 2, 3, 4, 5\}$ are $2, 3, 5$. $P(X \in \{2, 3, 5\}) = P(X=2) + P(X=3) + P(X=5)$ $= a \left( \frac{3}{4} + \frac{4}{8} + \frac{6}{32} ight) = a \left( \frac{24+16+6}{32} ight) = a \left( \frac{46}{32} ight) = a \left( \frac{23}{16} ight)$ $= \frac{4}{15} 	imes \frac{23}{16} = \frac{23}{60}$.

$X = x$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X = x)$	$0.15$	$0.23$	$k$	$0.10$	$0.20$	$0.08$	$0.07$	$0.05$

$x = x_{i}$	$0$	$1$	$2$
$P_{i}$	$\frac{25}{36}$	$\frac{5}{18}$	$\frac{1}{36}$

$X = x$	$1$	$2$	$3$	$4$	$5$	$6$
$P(X = x)$	$k$	$3k$	$5k$	$7k$	$8k$	$k$

$X=x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X=x)$	$0.05$	$0.1$	$2K$	$0$	$0.3$	$K$	$0.1$

If the probability function of a random variable $X$ is defined by $P(X=k) = a \left( \frac{k+1}{2^k} \right)$ for $k = 0, 1, 2, 3, 4, 5$,then the probability that $X$ takes a prime value is

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