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A football kick will result in a goal only if all of the following happen:
\[
\text{horizontal position is within } 3.5\text{ m of the centre of the goal,}
\]
\[
\text{the ball is below } 2.5\text{ m in height,}
\]
and
\[
\text{the goalkeeper does not save it.}
\]
The horizontal position of the kick is normally distributed with mean \(0\) and standard deviation \(3\) m:
\[
X \sim N(0,3^2).
\]
The height condition is satisfied with probability
\[
P(\text{under }2.5\text{ m}) = 0.7.
\]
If the ball is otherwise going in, the goalkeeper saves it with probability \(0.4\).
Find the probability that the kick results in a goal.
Check your answer
Solution
Answer: 31.8%
Let
\[
H = \{\text{kick is horizontally within }3.5\text{ m}\},
\]
\[
L = \{\text{kick is below }2.5\text{ m}\},
\]
\[
S = \{\text{goalkeeper saves the shot}\}.
\]
The kick scores only if it is horizontally on target, low enough, and not saved. So
\[
P(\text{goal}) = P(H)\,P(L)\,P(S^c).
\]
Since
\[
X \sim N(0,3^2),
\]
we want
\[
P(-3.5 \le X \le 3.5).
\]
Standardizing,
\[
Z=\frac{X-0}{3},
\]
so
\[
P(-3.5 \le X \le 3.5)
=
P\!\left(-\frac{3.5}{3} \le Z \le \frac{3.5}{3}\right).
\]
That is
\[
P(H)=P(-1.167 \le Z \le 1.167).
\]
Using the standard normal distribution:
\[
P(Z \le 1.167) \approx 0.8784.
\]
By symmetry,
\[
P(Z \le -1.167) \approx 0.1216.
\]
Hence
\[
P(H)=0.8784-0.1216=0.7568.
\]
Therefore, with 70% probability the kick is below 2.5m, and the goalkeeper not saving the shot with probability 60%:
\[
P(\text{goal}) = 0.7568 \times 0.7 \times 0.6
\]
\[
P(\text{goal}) \approx 0.318.
\]
So the probability of scoring is approximately 0.318.