In a box,there are 20 cards,out of which 10 a | vedclass

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(A) Let $P(A) = \frac{10}{20} = \frac{1}{2}$ and $P(B) = \frac{10}{20} = \frac{1}{2}$.We want the second $A$ to appear before the third $B$.This means

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$\frac{11}{16}$

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(A) Let $P(A) = \frac{10}{20} = \frac{1}{2}$ and $P(B) = \frac{10}{20} = \frac{1}{2}$.We want the second $A$ to appear before the third $B$.This means in the sequence of draws,we must have at most two $B$'s before the second $A$.The possible favorable sequences are:$1$. $AA$: Probability $= (\frac{1}{2})^2 = \frac{1}{4}$$2$. $ABA, BAA$: Probability $= 2 	imes (\frac{1}{2})^3 = \frac{2}{8} = \frac{1}{4}$$3$. $ABBA, BABA, BBAA$: Probability $= 3 	imes (\frac{1}{2})^4 = \frac{3}{16}$Summing these probabilities: $\frac{1}{4} + \frac{1}{4} + \frac{3}{16} = \frac{4}{16} + \frac{4}{16} + \frac{3}{16} = \frac{11}{16}$.

$X = x_{i}$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X = x_{i})$	$0.1$	$k$	$0.2$	$2k$	$3k$	$k$

In a box,there are $20$ cards,out of which $10$ are labelled as $A$ and the remaining $10$ are labelled as $B$. Cards are drawn at random,one after the other and with replacement,until a second $A$-card is obtained. The probability that the second $A$-card appears before the third $B$-card is

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