Solutions - 5090-04

(a)

$$\begin{array}{c}P[X\ge 8]=P[X=8]+P[X=9]+P[X=10]\\ \\ \gg \gg \left(\genfrac{}{}{0px}{}{10}{8}\right){(0.55)}^8{(0.45)}^2+\left(\genfrac{}{}{0px}{}{10}{9}\right){(0.55)}^9{(0.45)}^1+{(0.55)}^{10}{(0.45)}^0\\ \\ \gg \gg 0.0996\end{array}$$

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(b)

(1) Odds that a given team advances:

Let $A$ be the event that a given team (both members score $\ge 8$) advances. By independence:

$$P[A]={(0.0996)}^2$$

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(2) Compute probability “at least one team advances”:

The high school enters 2 independent teams.

$$\begin{array}{c}P[\text{at least one team advances}]=1-P{[A^c]}^2\\ \\ \gg \gg 1-{(1-{(0.0996)}^2)}^2\gg \gg 0.0197\end{array}$$