A six-faced die is biased such that 3 times P | vedclass

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(D) Let $P(	ext{prime}) = P(\{2, 3, 5\}) = 3p_1$,$P(	ext{composite}) = P(\{4, 6\}) = 2p_2$,and $P(1) = p_3$. Given $3(3p_1) = 6(2p_2) = 2p_3 = C$. T

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$\frac{8}{11}$

Vedclass · Answer

(D) Let $P(	ext{prime}) = P(\{2, 3, 5\}) = 3p_1$,$P(	ext{composite}) = P(\{4, 6\}) = 2p_2$,and $P(1) = p_3$. Given $3(3p_1) = 6(2p_2) = 2p_3 = C$. Then $P(	ext{prime}) = \frac{C}{3}$,$P(	ext{composite}) = \frac{C}{6}$,and $P(1) = \frac{C}{2}$. Since the sum of probabilities is $1$,we have $\frac{C}{3} + \frac{C}{6} + \frac{C}{2} = 1$,which gives $C = 1$. Thus,$P(	ext{prime}) = \frac{1}{3}$,$P(	ext{composite}) = \frac{1}{6}$,and $P(1) = \frac{1}{2}$. $A$ perfect square on a die is ${1, 4}$. $P(	ext{success}) = P(1) + P(4) = \frac{1}{2} + \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$. Wait,re-evaluating the given condition: $3 	imes P(	ext{prime}) = 6 	imes P(	ext{composite}) = 2 	imes P(1) = K$. $P(	ext{prime}) = K/3$,$P(	ext{composite}) = K/6$,$P(1) = K/2$. Sum: $K/3 + K/6 + K/2 = 1 \Rightarrow K = 1$. $P(	ext{prime}) = 1/3$,$P(	ext{composite}) = 1/6$,$P(1) = 1/2$. $P(	ext{perfect square}) = P(1) + P(4) = 1/2 + 1/6 = 4/6 = 2/3$. Mean of binomial distribution $X \sim B(n, p)$ is $np = 2 	imes (2/3) = 4/3$. Re-checking the provided solution logic: $P(	ext{prime}) = 2k$,$P(	ext{composite}) = k$,$P(1) = 3k$. $3(2k) = 6(k) = 2(3k) = 6k$. This matches the condition. Sum: $3(2k) + 2(k) + 3k = 1 \Rightarrow 6k + 2k + 3k = 11k = 1 \Rightarrow k = 1/11$. $P(	ext{success}) = P(1) + P(4) = 3k + k = 4k = 4/11$. Mean $= np = 2 	imes (4/11) = 8/11$.

$X = x$	$1$	$2$	$3$	$\dots$	$n$
$P(X = x)$	$\frac{1}{n}$	$\frac{1}{n}$	$\frac{1}{n}$	$\dots$	$\frac{1}{n}$

$X=x$	$1$	$2$	$3$	$4$
$P(X=x)$	$\frac{2}{20}$	$\frac{4}{20}$	$\frac{6}{20}$	$\frac{8}{20}$

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

$A$ six-faced die is biased such that $3 \times P(\text{a prime number}) = 6 \times P(\text{a composite number}) = 2 \times P(1)$. Let $X$ be a random variable that counts the number of times one gets a perfect square on some throws of this die. If the die is thrown twice,then the mean of $X$ is.

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