# Thread: Help with probability problems!...

1. ## Help with probability problems!...

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

2. Hello, Sneaky!

2. Suppose we are dealt five cards from a standard 52-card deck.

There are: .$\displaystyle \displaystyle _{52}C_5 \;=\;{52\choose5} \;=\;\frac{52!}{5!\,47!} \;=\;2,598,960\text{ possible Poker hands.}$

What is the probability that

a) we get all 4 Aces and the King of Spades

There is one way to get the four Aces $\displaystyle \{A\heartsuit,\:A\spadesuit,\:A\diamondsuit,\:A\cl ubsuit\}$
. . and one way to get the $\displaystyle K\spadesuit.$

$\displaystyle \text{Therefore: }\;P(\text{4 Aces} \wedge K\spadesuit) \;=\;\dfrac{1}{2,\!598,\!960}$

b) all 5 are Spades

There are: .$\displaystyle \displaystyle {13\choose5} \:=\:1287\text{ ways to get 5 Spades.}$

$\displaystyle \text{Therefore: }\:P(\text{5 Spades}) \;=\;\dfrac{1287}{2,\!598,\!960} \;=\;\dfrac{33}{66,\!640}$

c) We get no pairs (all are different values)

The 1st card can be any card: .$\displaystyle 52\text{ choices.}$

The 2nd card must not match the 1st card: .$\displaystyle 48\text{ choices.}$

The 3rd card must not match the 1st or 2nd card: .$\displaystyle 44\text{ choices.}$

The 4th card must not match the first 3 cards: .$\displaystyle 40\text{ choices.}$

The 5th card must not match the first 4 cards: .$\displaystyle 36\text{ choices.}$

But we have imparted an order to the five cards.

Since their order does not matter, there are:

. . $\displaystyle \dfrac{52\cdot48\cdot44\cdot40\cdot36}{5!} \;=\; 1,\!317,\!888 \text{ hands with no pairs.}$

$\displaystyle \text{Therefore: }\;P(\text{no pairs}) \;=\;\dfrac{1,\!317,\!888}{2,\!598,\!960} \;=\;\dfrac{2112}{4165}$

d) a Full House (3 of a kind and 2 of a kind)

There are 13 choices for the value of the Triple.
. . There are: .$\displaystyle {4\choose3} = 4\text{ ways to get the Triple.}$

There are 12 choices for the value of the Pair.
. . There are: .$\displaystyle {4\choose2} = 6\text{ ways to get the Pair.}$

Hence, there are: .$\displaystyle 13\cdot4\cdot12\cdot6 \:=\:3744\text{ Full Houses.}$

$\displaystyle \text{Therefore: }\;P(\text{Full House}) \;=\;\dfrac{3744}{2,\!598,\!960} \;=\;\dfrac{6}{4165}$

3. How can i complete the first question?