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Cards are drawn one-by-one without replacement from a well shuffled pack of 52 cards. Then, the probability that a face card (jack, queen or king) will appear for the first time on the third turn is equal to
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The correct answer is:
$\frac{12}{85}$
Ist turn Total number of face card $=12$
Total number of elements in sample space, $n(s)=52$
$\therefore P_{1}$ (no face card in first turn) $=\frac{52-12}{52}=\frac{40}{52}$
IInd turn $P_{2}$ (no face card in second turn)
$$
=\frac{(52-13)}{(52-1)}=\frac{39}{51}
$$
IIIrd turn $P_{3}$ (face card in third turn)
$$
=\frac{(52-40)}{(51-1)}=\frac{12}{50}
$$
$\therefore$ Required $P$ (face card on third turn)
$$
\begin{array}{l}
=P_{1} \times P_{2} \times P_{3} \\
=\frac{40}{52} \times \frac{39}{51} \times \frac{12}{50}=\frac{12}{85}
\end{array}
$$
Total number of elements in sample space, $n(s)=52$
$\therefore P_{1}$ (no face card in first turn) $=\frac{52-12}{52}=\frac{40}{52}$
IInd turn $P_{2}$ (no face card in second turn)
$$
=\frac{(52-13)}{(52-1)}=\frac{39}{51}
$$
IIIrd turn $P_{3}$ (face card in third turn)
$$
=\frac{(52-40)}{(51-1)}=\frac{12}{50}
$$
$\therefore$ Required $P$ (face card on third turn)
$$
\begin{array}{l}
=P_{1} \times P_{2} \times P_{3} \\
=\frac{40}{52} \times \frac{39}{51} \times \frac{12}{50}=\frac{12}{85}
\end{array}
$$
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