A card is drawn from a pack of 52 playing car | vedclass

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Question

(C) The probability of drawing a heart is $p_1 = \frac{13}{52} = \frac{1}{4}$.The probability of drawing a diamond is $p_2 = \frac{13}{52} = \frac{1}{

Vedclass · Accepted Answer

$90 	imes (\frac{1}{2})^{10}$

Vedclass · Answer

(C) The probability of drawing a heart is $p_1 = \frac{13}{52} = \frac{1}{4}$.The probability of drawing a diamond is $p_2 = \frac{13}{52} = \frac{1}{4}$.The probability of drawing a black card is $p_3 = \frac{26}{52} = \frac{1}{2}$.Since the cards are replaced,we use the multinomial distribution formula for $n=6$ trials with outcomes $n_1=2, n_2=2, n_3=2$:$P = \frac{n!}{n_1! n_2! n_3!} 	imes (p_1)^{n_1} 	imes (p_2)^{n_2} 	imes (p_3)^{n_3}$$P = \frac{6!}{2! 2! 2!} 	imes (\frac{1}{4})^2 	imes (\frac{1}{4})^2 	imes (\frac{1}{2})^2$$P = \frac{720}{8} 	imes \frac{1}{16} 	imes \frac{1}{16} 	imes \frac{1}{4}$$P = 90 	imes \frac{1}{4^2} 	imes \frac{1}{4^2} 	imes \frac{1}{2^2} = 90 	imes \frac{1}{2^4} 	imes \frac{1}{2^4} 	imes \frac{1}{2^2} = 90 	imes (\frac{1}{2})^{10}$

$X = x$	$0$	$1$	$2$
$P(X = x)$	$4k - 10k^2$	$5k - 1$	$3k^3$

$X=x$	$1$	$2$	$3$	$4$
$P(X=x)$	$2k$	$4k$	$3k$	$k$

$X=k$	$0$	$1$	$2$	$3$	$4$
$P(X=k)$	$0.1$	$0.4$	$0.3$	$0.2$	$0$

$A$ card is drawn from a pack of $52$ playing cards. The card is replaced and the pack is shuffled. If this process is repeated $6$ times,what is the probability that $2$ hearts,$2$ diamonds,and $2$ black cards are drawn?

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