# Four cards are drawn one by one with replacement from a well-shuffled deck of playing cards. Find the probability that at least three cards are of diamonds.

Here  probability of getting diamond $p= \frac{13}{53}= \frac{1}{4}$
$q= \frac{3}{4}$
$\therefore$ p(at least 3 are diamonds) $= p\left ( x= 3 \right )+p\left ( x= 4 \right )$
$=\, ^{4}C_{3}\left ( \frac{1}{4} \right )^{3}\left ( \frac{3}{4} \right )^{4-3}+\, ^{4}C_{4}\left ( \frac{1}{4} \right )^{4}\left ( \frac{3}{4} \right )^{4-4}$
$\Rightarrow \frac{12}{256}+\frac{1}{256}= \frac{13}{256}$

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