sneaky666
Sep22-10, 11:18 PM
1. The problem statement, all variables and given/known data
1. Suppose we roll 10 fair 6-sided dice. What is the probability that there are exactly two 2's showing?
2. Suppose we are dealt five cards from a standard 52-card deck. What is the probability that
a) we get all 4 aces and the king of spades
b) all 5 are spades
c) we get no pairs (all are different values)
d) a full house (3 of a kind and 2 of a kind)
2. Relevant equations
This one is correct:
there are 2 pots
in pot1 there is 5 red balls and 7 blue balls
in pot2 there is 6 red balls and 12 blue balls
3 balls are chosen randomely from each pot
chances of all 6 balls to be same color = P(A)
chances of all 6 balls to be red = P(B)
chances of all 6 balls to be blue = P(C)
P(A) = P(B or C)
=P(B) + P(C) -0
P(B)=|B|/|S| = |B|/(12choose3)(18choose3) =
(5choose3)(7choose0)(6choose3)(12choose0)/(12choose3)(18choose3) = 5/4488
P(C)=|C|/|S| = |C|/(12choose3)(18choose3) =
(5choose0)(7choose3)(6choose0)(12choose3)/(12choose3)(18choose3) = 35/816
P(A) = 5/4488 + 35/816 - 0 = 395/8976
3. The attempt at a solution
1.
number of outcomes = 6^10 = 60466176
10!/2!8! = 45
so i get
45/60466176
2.
number of outcomes = 52x51x50x49x48 / 5x4x3x2x1 = 2598960
a) Here i have two different methods, i don't know if both are wrong or one is right...
method 1
(4choose1 * 4choose4 ) / 2598960 = 1/649740
method 2
( (13choose1) * (13choose1) * (13choose1) * (13choose1) ) / 2598960 = ~0.066
b)
(13choose5)/2598960 = ~4.95x10^(-4)
c)
( (4choose1)*(4choose1)*(4choose1)*(4choose1)*(4choo se1) ) /2598960 = ~3.95x10^(-4)
d)
(4choose3)*(4choose2) /2598960 = 1/108290
1. Suppose we roll 10 fair 6-sided dice. What is the probability that there are exactly two 2's showing?
2. Suppose we are dealt five cards from a standard 52-card deck. What is the probability that
a) we get all 4 aces and the king of spades
b) all 5 are spades
c) we get no pairs (all are different values)
d) a full house (3 of a kind and 2 of a kind)
2. Relevant equations
This one is correct:
there are 2 pots
in pot1 there is 5 red balls and 7 blue balls
in pot2 there is 6 red balls and 12 blue balls
3 balls are chosen randomely from each pot
chances of all 6 balls to be same color = P(A)
chances of all 6 balls to be red = P(B)
chances of all 6 balls to be blue = P(C)
P(A) = P(B or C)
=P(B) + P(C) -0
P(B)=|B|/|S| = |B|/(12choose3)(18choose3) =
(5choose3)(7choose0)(6choose3)(12choose0)/(12choose3)(18choose3) = 5/4488
P(C)=|C|/|S| = |C|/(12choose3)(18choose3) =
(5choose0)(7choose3)(6choose0)(12choose3)/(12choose3)(18choose3) = 35/816
P(A) = 5/4488 + 35/816 - 0 = 395/8976
3. The attempt at a solution
1.
number of outcomes = 6^10 = 60466176
10!/2!8! = 45
so i get
45/60466176
2.
number of outcomes = 52x51x50x49x48 / 5x4x3x2x1 = 2598960
a) Here i have two different methods, i don't know if both are wrong or one is right...
method 1
(4choose1 * 4choose4 ) / 2598960 = 1/649740
method 2
( (13choose1) * (13choose1) * (13choose1) * (13choose1) ) / 2598960 = ~0.066
b)
(13choose5)/2598960 = ~4.95x10^(-4)
c)
( (4choose1)*(4choose1)*(4choose1)*(4choose1)*(4choo se1) ) /2598960 = ~3.95x10^(-4)
d)
(4choose3)*(4choose2) /2598960 = 1/108290