# The experimental probability that Kristen will hit the ball when she is at bat is 3/5. If she is at bat 80 times in a season, how many times can Kristen expect to hit the ball?

Feb 16, 2018

48 times

#### Explanation:

Number of times she is expected to hit the ball

$= P \times \text{Total times she bat}$

$= \frac{3}{5} \times 80$

$= \frac{3}{\cancel{5}} \times {\cancel{80}}^{16}$

$= 3 \times 16$

$= 48$ times

Feb 16, 2018

$48 \text{ times}$

#### Explanation:

$\text{We can just do "(3/5)*80 = 48". If you want a proof then}$
$\text{read further here underneath.}$

$P \left[\text{Kristen hits k times on 80}\right] = C \left(80 , k\right) {\left(\frac{3}{5}\right)}^{k} {\left(\frac{2}{5}\right)}^{80 - k}$
$\text{with "C(n,k) = (n!)/((n-k)!*(k!)) " (combinations)}$
$\text{(binomial distribution)}$

$\text{Expected value = average = E[k] :}$

${\sum}_{k = 0}^{k = 80} k \cdot C \left(80 , k\right) {\left(\frac{3}{5}\right)}^{k} {\left(\frac{2}{5}\right)}^{80 - k}$
= sum_{k=1}^{k=80} 80*(79!)/((80-k)! (k-1)!) (3/5)^k (2/5)^(80-k)
$= 80 \cdot \left(\frac{3}{5}\right) {\sum}_{k = 1}^{k = 80} C \left(79 , k - 1\right) {\left(\frac{3}{5}\right)}^{k - 1} {\left(\frac{2}{5}\right)}^{80 - k}$
$= 80 \cdot \left(\frac{3}{5}\right) {\sum}_{t = 0}^{t = 79} C \left(79 , t\right) {\left(\frac{3}{5}\right)}^{t} {\left(\frac{2}{5}\right)}^{79 - t}$
$\text{(with "t=k-1")}$
$= 80 \cdot \left(\frac{3}{5}\right) \cdot 1$
$= 48$

$\text{So for a binomial experiment, with "n" tries, and probability}$
$p \text{ for the chance of success on a single try, we have in general}$
$\text{expected value=average= "n*p" (of the number of successes)}$