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One ticket is selected at random from 50 tickets numbered $00,01,02, \ldots \ldots 49$. The probability that sum of the digits is 10 , given that product of the digits is 9 , is
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The correct answer is:
$1 / 2$
$A=$ Sum of digits is 10
$B=$ Product of digits is 9
$A=\{19,28,37,46\}$
$\Rightarrow n(A)=4$
$n(B)=2$
$n(S)=50$
$B=\{19,33\}$
$P(A)=\frac{4}{50}$ and $P(B)=\frac{2}{50}$
$P(A \cap B)=\frac{1}{50}$
Required probability, $P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}=\frac{1}{2}$
$B=$ Product of digits is 9
$A=\{19,28,37,46\}$
$\Rightarrow n(A)=4$
$n(B)=2$
$n(S)=50$
$B=\{19,33\}$
$P(A)=\frac{4}{50}$ and $P(B)=\frac{2}{50}$
$P(A \cap B)=\frac{1}{50}$
Required probability, $P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}=\frac{1}{2}$
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