# Hypergeometric Distribution

• Oct 7th 2009, 10:43 AM
Walcott89
Hypergeometric Distribution
I would greatly appreciate if someone could help me set up and solve the following two problems.

1. If 7 cards are dealt from an ordinary deck of 52 cards, what is the probability that exactly 2 of them will be face cards? That at least 1 of them will be a queen?

Using the hypergeometric distribution formula, I said that this would be equal to 12 choose 3 divided by 52 choose 7. What am I doing wrong?

2. A random committee of size 3 is selected from 4 doctors and 2 nurses. Write a formula for the probability distribution of the random variable X representing the number of doctors on the committee. Find P(2<=X<=3).

Thanks! (Happy)
• Oct 7th 2009, 12:38 PM
Soroban
Hello, Walcott89!

Quote:

1. 7 cards are dealt from an ordinary deck of 52 cards.
(a) What is the probability that exactly 2 of them will be face cards?

There are: .$\displaystyle {52\choose7}$ possible outcomes.

There are: .12 Face Cards and 40 Others.

We want 2 Face Cards and 5 Others.
. . There are: .$\displaystyle {12\choose2}{40\choose5}$ ways.

Therefore: .$\displaystyle P(\text{exactly 2 Face Cards}) \;=\;\dfrac{{12\choose2}{40\choose5}}{{52\choose7} }$

Quote:

[(b) What is the probability that at least 1 of them will be a Queen?
There are: 4 Queens and 48 Others.

Find the probability of no Queens.
Then we want 7 Others.
. . There are: .$\displaystyle {48\choose7}$ ways.
Hence: .$\displaystyle P(\text{no Queens}) \;=\;\frac{{48\choose7}}{{52\choose7}}$
Therefore: .$\displaystyle P(\text{at least one Queen}) \;\;=\;\;1 - \frac{{48\choose7}}{{52\choose7}}$

• Oct 10th 2009, 09:35 AM
qpmathelp
Quote:

Originally Posted by Walcott89
I would greatly appreciate if someone could help me set up and solve the following two problems.

1. If 7 cards are dealt from an ordinary deck of 52 cards, what is the probability that exactly 2 of them will be face cards? That at least 1 of them will be a queen?

Using the hypergeometric distribution formula, I said that this would be equal to 12 choose 3 divided by 52 choose 7. What am I doing wrong?

2. A random committee of size 3 is selected from 4 doctors and 2 nurses. Write a formula for the probability distribution of the random variable X representing the number of doctors on the committee. Find P(2<=X<=3).

Thanks! (Happy)

2.

you have to choose x doctors and (3-x) nurses in the sample of 3

C(3,x) C[2,(3-x)] / C(6,3)

binomial distributionmixture: Binomial distribution