PROBABILITY

G

Guest

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.
 
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.

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