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Probability problem
- Thread starter Kylopod
- Start date
Well, try here: http://www.mathgoodies.com/lessons/vol6 ... ional.html
Google will give you more sites.
Google will give you more sites.
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?
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.
I suspect that the numbers are incorrect.
We are told that 40% of the 56 people are fat.
. . \(\displaystyle 0.40 \times 56 \:=\:22.4\)
\(\displaystyle \text{How can we have }22\tfrac{2}{5}\text{ fat people?}\)
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.
. . \(\displaystyle 0.40 \times 56 \:=\:22.4\)
\(\displaystyle \text{How can we have }22\tfrac{2}{5}\text{ fat people?}\)