Question 1194848: In a 5-card poker hand, find the probability that the hand contains 3 spades.
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
There are 2,598,960 different five-card hands possible (whether they have 3 spades or not).
The scratch work for this is shown here
n C r = (n!)/(r!(n-r)!)
52 C 5 = (52!)/(5!*(52-5)!)
52 C 5 = (52!)/(5!*47!)
52 C 5 = (52*51*50*49*48*47!)/(5!*47!)
52 C 5 = (52*51*50*49*48)/(5!)
52 C 5 = (52*51*50*49*48)/(5*4*3*2*1)
52 C 5 = (311875200)/(120)
52 C 5 = 2,598,960
We use the nCr combination formula because order doesn't matter in a card hand.
A hand like {Q, 2, 3, J, K} is the same as {2, 3, J, Q, K}
There are 13 spades and we want exactly 3 spades
Let's plug in n = 13 and r = 3
n C r = (n!)/(r!(n-r)!)
13 C 3 = (13!)/(3!*(13-3)!)
13 C 3 = (13!)/(3!*10!)
13 C 3 = (13*12*11*10!)/(3!*10!)
13 C 3 = (13*12*11)/(3!)
13 C 3 = (13*12*11)/(3*2*1)
13 C 3 = (1716)/(6)
13 C 3 = 286
There are 286 ways to pick those three spades where order doesn't matter.
Then there are 39C2 = 741 ways to pick the other two non-spade cards. I'll skip the steps in showing how to get the 741, but you'd follow the templates above to show your teacher how you got 741.
We have 286*741 = 211,926 ways to pick a five-card hand that has exactly three spades and two other non-spade cards.
Divide the results we found
A = number of hands that have 3 spades + 2 others
B = number of five card hands total
A = 211,926
B = 2,598,960
A/B = probability of getting exactly 3 spades and 2 other non-spades
A/B = (211,926)/(2,598,960)
A/B = 0.081543
Normally I would stick to the fraction form, but the integer values are quite large.
It might make more intuitive sense to go for the approximate decimal form.
Of course it's best to follow whatever instructions your teacher provides when it comes to recording the final answer.
Answer as a fraction: (211,926)/(2,598,960)
Answer in decimal form: 0.081543 which is approximate
Answer in percent form: 8.1543% which is also approximate
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