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14.Probability
normal
A card is drawn from a pack of $52$ playing cards. The card is replaced and pack is shuffled. If this is done six times, then the probability that $2$ hearts, $2$ diamond and $2$ black cards are drawn is
A
$90 \times (\frac{1}{4})^6$
B
$\frac{45}{2} (\frac{3}{4})^4 $
C
$90 \times (\frac{1}{2})^{10} $
D
$(\frac{1}{2})^{10}$
Solution
$ P =\frac{13_{c_{1}}}{52_{c_{1}}} \times \frac{13_{c_{1}}}{52_{c_{1}}} \times \frac{13_{c_{1}}}{52_{c_{1}}} \times \frac{13_{c_{1}}}{52_{c_{1}}} \times \frac{26_{c_{1}}}{52_{c_{1}}} \times \frac{26_{c_{1}}}{52_{c_{1}}} \times \frac{6 !}{2 ! 2 ! 2 !} $
$=90 \times\left(\frac{1}{2}\right)^{10} $
Standard 11
Mathematics
