# Binomial probabilities. In a city, the percentage of left-handed women is 16% and the percentage of left-handed men is 22%. A random sample of five men and five women are selected from the population of the city. What is the probability that the sample contains at least one left-handed woman and one left-handed man.

Binomial probabilities
In a city, the percentage of left-handed women is 16% and the percentage of left-handed men is 22%. A random sample of five men and five women are selected from the population of the city. What is the probability that the sample contains at least one left-handed woman and one left-handed man.
So yeah, using my awesome calculator I found probabilities of 0 men - 0.2289. 0 women - 0.418. It follows therefore that at least one man - 0.711, one woman - 0.582. My intuition told me that I now have to multiply last two values and even got me the right answer. Could anyone please explain why we multiply 0.711 and 0.582?
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Miya Swanson
Step 1
You multiply them because you are looking for the probability of at least one left-handed woman AND at least one left-handed man.
Step 2
If you let A be the event that the sample contains at least 1 left-handed woman and let B be the event that the sample contains at least 1 left-handed man, then $Pr\left(A\cap B\right)=Pr\left(A\right)×Pr\left(B\right)$