A six-faced fair die is thrown until 2 appear | vedclass

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(D) Let $p$ be the probability of getting a $2$ in a single throw,so $p = \frac{1}{6}$.Let $q$ be the probability of not getting a $2$,so $q = 1 - p =

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$\frac{5}{11}$

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(D) Let $p$ be the probability of getting a $2$ in a single throw,so $p = \frac{1}{6}$.Let $q$ be the probability of not getting a $2$,so $q = 1 - p = \frac{5}{6}$.The event that $2$ appears in an even number of trials means it appears on the $2^{nd}, 4^{th}, 6^{th}, \dots$ trial.This corresponds to the sequences: (not $2$,$2$) or (not $2$,not $2$,not $2$,$2$) or $\dots$The probability is given by the infinite geometric series:$P = qp + q^3p + q^5p + \dots$This is a geometric series with first term $a = qp = \frac{5}{6} 	imes \frac{1}{6} = \frac{5}{36}$ and common ratio $r = q^2 = (\frac{5}{6})^2 = \frac{25}{36}$.The sum of an infinite geometric series is $S = \frac{a}{1-r}$.$P = \frac{\frac{5}{36}}{1 - \frac{25}{36}} = \frac{\frac{5}{36}}{\frac{11}{36}} = \frac{5}{11}$.

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

$X = x$	$-4$	$-2$	$0$	$2$	$4$	$6$	$8$	$10$
$F(X = x)$	$0.1$	$0.3$	$0.5$	$0.65$	$0.75$	$0.85$	$0.90$	$1$

$A$ six-faced fair die is thrown until $2$ appears. What is the probability that $2$ appears in an even number of trials? (The die has six faces numbered $1, 2, 3, 4, 5,$ and $6$.)

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