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

Vedclass · Accepted Answer

$\frac{2}{19}$

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(B) The total number of tickets is $100$ (from $00$ to $99$).Let $Y$ be the product of the digits. The event $(Y = 0)$ occurs if at least one of the digits is $0$.The tickets where at least one digit is $0$ are: ${00, 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 20, 30, 40, 50, 60, 70, 80, 90}$.Counting these,there are $10$ tickets starting with $0$ ($00$ to $09$) and $9$ tickets ending with $0$ $(10, 20, \dots, 90)$.Thus,the number of favorable outcomes for $(Y = 0)$ is $10 + 9 = 19$.So,$P(Y = 0) = \frac{19}{100}$.Now,let $X$ be the sum of the digits. We need to find $(X = 9) \cap (Y = 0)$.This means the sum of the digits is $9$ $AND$ at least one digit is $0$.The tickets satisfying this are $09$ $(0+9=9)$ and $90$ $(9+0=9)$.So,the number of favorable outcomes for $(X = 9) \cap (Y = 0)$ is $2$.Thus,$P(X = 9 \cap Y = 0) = \frac{2}{100}$.The conditional probability is $P(X = 9 | Y = 0) = \frac{P(X = 9 \cap Y = 0)}{P(Y = 0)} = \frac{2/100}{19/100} = \frac{2}{19}$.

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,then $P(X = 9 | Y = 0)$ equals

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