Asked by Candice

Hi, could you please help me? I have been stick on this question for a week and I have tried getting help from my teacher and parents, but I still cant get it. TIA

A standard deck of 52 cards consists of 4 suits (hearts, clubs, diamonds, and spades). If a student deals 3 cards from a standard deck, determine the probability to the nearest hundredth, that at least one card is a heart. Use the method stated below:

A.) Direct Reasoning (2 marks)
B.) Indirect Reasoning (2 marks)
Answered by Damon
52/4 = 13 hearts in the deck
52-13 = 39 are not hearts

probability that all 3 are H
= 13/52 *12/51*11/50 = .01294

probability that 2/3 are H
=13/52 * 12/51* 39/50 = .04588
{39 is number of non hearts}

probability that 1/3 is H
= 13/52 *39/51*38/50 = .14529
add
A) 0.2041
Answered by Damon
call h a heart
call x another suit
d = 52*51*50 = 132600
combos we could get
hhh =13*12*11/d = 1716/d
hhx =13*12*39/d = 6084/d
hxx =13*39*38/d = 19266/d
hxh =13*39*12/d = 6084/d
xxh =39*38*13/d = 19266/d
xhx =39*13*38/d = 19266/d
xhh =39*13*12/d = 6084/d
add to get
77766/d= 77766/132600 = .58647
That is the hard way
now
to get ALL x, no hearts
39/52*38/51*37/50 = 54834/d =.41353
1 - .41353 = .58647 !!! whew, the easy way
Answered by Damon
When I did it the easy way, I got a different answer, so went back and plodded through the hard way.
Answered by Steve
P(some hearts) = 1-P(no hearts)
= 1 - (39/52)(38/51)(37/50) = 0.58647
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