# Thread: Bridge game distribution of cards

1. ## Bridge game distribution of cards

63. you are dealt a 13-card bridge hand from a 52-card bridge deck.
b. what is the probability you will be dealt at least 2 hearts?

so i think i figured this out... just not sure if i'm missing more to the equation...

N=52 (population)
n=13(sample size)
x=2(number of successes)
number of failures = n-x
= 13-2
= 11
P(at least 2 hearts) = 13 39
2 11
---------
52
13
= 78(1,675,986,051) / 6.53 x 10^11
= 1.307 x 10^11 / 6.35 x 10^11
= 0.20583
therefore, the probability you will receive at least 2 hearts is 0.20583 or 20.583% chance.

c. What is the probability that you will receive 5 hears, 4 clubs, 2 diamonds and 1 spade?

so i already calculated out for 5 hearts in part a. and came to 0.12472

N=52 still?
n=8 (we already took 5 cards)
x= 4
number of failures = n-x
= 8-4
= 6

P(4 clubs) = 8 39
4 6
---------
52
8
= 70(3,262,623)/752,538,150
= 228,383,610/752,538,150
= 0.30345

now N=52
n = 4
x = 2
number of failures = n-x
= 4-2
= 2
P(2 diamonds) = 4 39
2 2
---------
52
4
= 6(741)/270,725
= 4,446/ 270,725
= 0.01642
N=52
n=2
x=1
number of failures=n-x
= 2-1
= 1
P(1 club) = 2 39
1 1
---------
52
2
= 2(39)/1,326
= 78/1,326
= 0.05882

then to find the total probability it would be;

Pt = P(5 hearts) + P(4 spades) + P(2 diamonds) + P(1 club)
= 0.12472 + 0.30345 + 0.01642 + 0.05882
= 0.50341

Therefore the probability that you will be dealt 5 hearts, 4 spades, 2 diamonds and 1 club is 0.50341 or 50.341%?

Again with this question i just want to make sure that i've got this figured out the right way...

2. I cant really understand the working you have used for (b) but this is how I have done it;
I think it is right..
$P(\mbox{2 or more hearts})=1-[P(\mbox{0 hearts})+P(\mbox{1 heart})]$
$
P(\mbox{0 hearts})=\frac{13C0\cdot 39C13}{52C13}=\frac{ \frac{13!}{0!13!}\cdot \frac{39!}{13!\cdot 26!} } {\frac {52!}{13! \cdot 39!}}=0.012791$

$
P(\mbox{1 hearts})=\frac{13C1\cdot 38C12}{52C13}=\frac{ \frac{13!}{1!12!}\cdot \frac{38!}{12!\cdot 26!} } {\frac {52!}{13! \cdot 39!}}=0.055427$

so
$P(\mbox{2 or more hearts})=1-(0.012791+0.055427)=0.93178161$
So 93% chance of 2 or more hearts.

A similiar method can be used for part (c), however, you have only listed 12 cards being delt?