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Answer 1 · 2016-07-06T01:34:10

$= 6 C_{4} {(\frac{1}{4})}^{4} {(\frac{3}{4})}^{2}$ $=6C_4(1/4)^4(3/4)^2$

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

There are 13 hearts in the pack of 52..

The probability of drawing a heart

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

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

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,

$= 6 C_{4} {(\frac{1}{4})}^{4} {(\frac{3}{4})}^{2}$ $=6C_4(1/4)^4(3/4)^2$

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