A coin is biased so that the head is 3 times | vedclass

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(A) Given that the probability of head $P(H) = 3P(T)$. Since $P(H) + P(T) = 1$,we have $4P(T) = 1$,so $P(T) = \frac{1}{4}$ and $P(H) = \frac{3}{4}$.Th

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$\frac{21}{16}$

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(A) Given that the probability of head $P(H) = 3P(T)$. Since $P(H) + P(T) = 1$,we have $4P(T) = 1$,so $P(T) = \frac{1}{4}$ and $P(H) = \frac{3}{4}$.The coin is tossed until a head appears or three tails appear. The possible values for $X$ are $1, 2, 3$.For $X=1$: The outcome is $H$. $P(X=1) = P(H) = \frac{3}{4}$.For $X=2$: The outcome is $TH$. $P(X=2) = P(T) 	imes P(H) = \frac{1}{4} 	imes \frac{3}{4} = \frac{3}{16}$.For $X=3$: The outcomes are $TTH$ or $TTT$. $P(X=3) = P(T)^2 	imes P(H) + P(T)^3 = (\frac{1}{4})^2 	imes \frac{3}{4} + (\frac{1}{4})^3 = \frac{3}{64} + \frac{1}{64} = \frac{4}{64} = \frac{1}{16}$.The mean $E(X) = \sum x_i P(x_i) = 1(\frac{3}{4}) + 2(\frac{3}{16}) + 3(\frac{1}{16}) = \frac{3}{4} + \frac{6}{16} + \frac{3}{16} = \frac{12}{16} + \frac{9}{16} = \frac{21}{16}$.

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

$X = x$	$0$	$2$	$4$	$6$	$8$	$10$
$P(X = x)$	$0$	$k$	$2k$	$5k^2$	$2k^2$	$3k$

$X = x$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$\frac{2}{10}$	$\frac{3}{10}$	$\frac{4}{10}$

$A$ coin is biased so that the head is $3$ times as likely to occur as tail. This coin is tossed until a head or three tails occur. If $X$ denotes the number of tosses of the coin,then the mean of $X$ is

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