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.

 

 

 

 
 
 
 
 

Answers (1)

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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