Question da498

Jul 6, 2016

$= 6 {C}_{4} {\left(\frac{1}{4}\right)}^{4} {\left(\frac{3}{4}\right)}^{2}$

$= 15 X \left(\frac{1}{256}\right) \left(\frac{81}{256}\right) = 0.01854 ,$ nearly..

Explanation:

There are 13 hearts in the pack of 52..

The probability of drawing a heart

= (number of favorable cases)/(total number of cases)

$= \frac{13}{52} = \frac{1}{4}$.

The probability of not drawing a heart =1-1/4=3/4.

So, for drawing exactly 4 hearts in six draws in succession,

replacing the card drawn in each draw, the probability is the

compound probability,

$= 6 {C}_{4} {\left(\frac{1}{4}\right)}^{4} {\left(\frac{3}{4}\right)}^{2}$

$= 15 X \left(\frac{1}{256}\right) \left(\frac{81}{256}\right) = 0.01854 ,$ nearly....