91Ó°ÊÓ

Girl A throws snowballs at a target 10 times and probability that she will hit the target is 0.3.

Girl B throws snowballs at a target 15 times and probability that she will hit the target is 0.2.

Girl B throws snowballs at a target 20 times and probability that she will hit the target is 0.1.

The distribution of the total number of times X that the target is hit will be approximately the normal distribution with mean

\(10\left( {0.3} \right) + 15\left( {0.20} \right) + 20\left( {0.1} \right) = 8\)

Variance is

\(10\left( {0.3} \right)\left( {0.7} \right) + 15\left( {0.2} \right)\left( {0.8} \right) + 20\left( {0.1} \right)\left( {0.8} \right) = 6.3\)

Therefore the distribution of X is approximately a standard normal distribution.

\(\begin{array}{c}Z = \frac{{\left( {X - 8} \right)}}{{\sqrt {6.3} }}\\ = \frac{{\left( {X - 8} \right)}}{{2.51}}\end{array}\)

Thus,

\(\begin{array}{c}{\rm P}\left( {X \ge 12} \right) = {\rm P}\left( {X \ge 11.5} \right)\\ = {\rm P}\left( {Z \ge \frac{{11.5 - 8}}{{2.51}}} \right)\\ = {\rm P}\left( {Z \ge \frac{{3.5}}{{2.51}}} \right)\end{array}\)

\(\begin{array}{l}{\rm P}\left( {X \ge 12} \right) \approx 1 - \phi \left( {1.394} \right)\\{\rm P}\left( {X \ge 12} \right) \approx 0.082\end{array}\)

Therefore,Probability that the target will be hit at least 12 times is 0.082.