Probability problem

Kylopod

New member
Joined
Feb 10, 2009
Messages
3
56 people are in a room. 21 are male and the rest female. 40% of the people are fat and the rest skinny. What is the probability that exactly three of the people (no more, no less) are fat males?
 
I didn't get anywhere, because I've forgotten how to do this kind of problem. 40% of 56 is 22.4.
 
I already understand that I can calculate the probable number of fat males as (21/56)*(0.4) = 15%, which comes out to 8.4.

My question, however, is how do I calculate the probability that there will be only 3 fat males--considerably less than predicted?
 
Hello, Kylopod!

Are you sure of the wording of the problem?
As written, it doesn't make sense.


56 people are in a room: 21 male and 35 female.
40% of the people are fat and the rest (60%) skinny.
What is the probability that exactly three of the people (no more, no less) are fat males?

I suspect that the numbers are incorrect.

We are told that 40% of the 56 people are fat.
. . \(\displaystyle 0.40 \times 56 \:=\:22.4\)

\(\displaystyle \text{How can we have }22\tfrac{2}{5}\text{ fat people?}\)

 
Top