- #1

sneaky666

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## Homework Statement

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)

## Homework 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

## 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)*(4choose1) ) /2598960 = ~3.95x10^(-4)

d)

(4choose3)*(4choose2) /2598960 = 1/108290