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Team A has $14$ females and $11$ males. Team B has $16$ females and $15$ males. The Teams are competing against each other. What is the probability a male and a female are chosen to compete?

Team $A$ selecting a male is $\frac{11}{25}$ $\times$ Team B selecting a female $\frac{16}{31}$.

Team A selecting a female is $\frac{14}{25}$ $\times$ Team B selecting a male is $\frac{15}{31}$. But how do we then combine the two different selections?

The Empty String Photographer
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Maths help
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    The probability of the union of two mutually exclusive events is the sum of the probabilities of these events. – drhab Mar 27 '23 at 07:48

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First, we need to multiply the fractions together: $$\frac{11}{25}\times\frac{16}{31}=\frac{176}{775}$$ $$\frac{14}{25}\times\frac{15}{31}=\frac{210}{775}$$ Then, to calculate the total probability of either event (as @drhab said), we add the fractions together: $$\frac{176}{775}+\frac{210}{775}=\frac{386}{775}$$ The fraction is in it's simplest form, so the answer is $\frac{386}{775}$.

The Empty String Photographer
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