PROBABILITY

[Guest]

We are drawing cards without replacement from the infamous "Iraq's Most Wanted" deck issued by the U.S. Military. If I was drawing from the full deck of 55 cards, what are the following probabilities:

a. I draw a card that is not Saddam Hussein

b. I draw three cards, which end up being Saddam Hussein and his two sons.

c. I draw 14 cards and not one of them is Saddam Hussein.

 

[soroban]

Hello, IN IRAQ!

We are drawing cards without replacement from the infamous "Iraq's Most Wanted" deck issued by the U.S. Military.
If I was drawing from the full deck of 55 cards, what are the following probabilities:

a. I draw a card that is not Saddam Hussein

There are 55 cards; 54 of them are not Saddam.

\(\displaystyle \;\;P(\text{not Saddam})\:=\:\frac{54}{55}\)

b. I draw three cards, which end up being Saddam Hussein and his two sons.

There are \(\displaystyle C(55,3)\,=\,\frac{55!}{3!\cdot52!}\,=\,26,235\) possible sets of 3-card hands.

There is one hand which contains Saddam and his two sons.

\(\displaystyle \;\;P(\text{Saddam & two sons}) \:=\:\frac{1}{26,235\)

c. I draw 14 cards and not one of them is Saddam Hussein.

There are \(\displaystyle C(55,14)\,=\,\frac{55!}{14!\cdot41!}\) possible 14-card hands.

There are 54 non-Saddam cards.
There are \(\displaystyle C(54,14)\,=\,\frac{54!}{14!\cdot40!}\) 14-card hands without Saddam.

\(\displaystyle \;\;P(\text{14 non-Saddams})\;=\;\L\frac{\frac{54!}{14!\cdot40!}}{\frac{55!}{14!\cdot41!}} \;= \;\frac{41}{55}\)