One ticket is selected at random from 100 tic | vedclass

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(B) The total number of tickets is $100$ (from $00$ to $99$).Let $Y = 0$ be the event that the product of the digits is $0$. This occurs if at least o

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$\frac{2}{19}$

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(B) The total number of tickets is $100$ (from $00$ to $99$).Let $Y = 0$ be the event that the product of the digits is $0$. This occurs if at least one digit is $0$. The tickets are: $00, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 20, 30, 40, 50, 60, 70, 80, 90$.Counting these,we have $10$ tickets starting with $0$ ($00$ to $09$) and $9$ tickets ending with $0$ $(10, 20, \dots, 90)$. Note that $00$ is counted in both,so the total number of distinct tickets is $10 + 9 = 19$.Thus,$P(Y = 0) = \frac{19}{100}$.Now,let $X = 9$ be the event that the sum of the digits is $9$. We look for the intersection $(X = 9) \cap (Y = 0)$,which means the sum of digits is $9$ $AND$ at least one digit is $0$. The only such tickets are $09$ and $90$.Thus,$P(X = 9 \cap Y = 0) = \frac{2}{100}$.The conditional probability is $P(X = 9 | Y = 0) = \frac{P(X = 9 \cap Y = 0)}{P(Y = 0)} = \frac{2/100}{19/100} = \frac{2}{19}$.

$Face$	$1$	$2$	$3$	$4$	$5$	$6$
$Probability$	$0.1$	$0.32$	$0.21$	$0.15$	$0.05$	$0.17$

One ticket is selected at random from $100$ tickets numbered $00, 01, 02, \dots, 98, 99$. If $X$ and $Y$ denote the sum and the product of the digits on the tickets respectively,then $P(X = 9 | Y = 0)$ equals:

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