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Q. 1

Expert-verifiedFound in: Page 667

Book edition
4th

Author(s)
David Moore,Daren Starnes,Dan Yates

Pages
809 pages

ISBN
9781319113339

Suppose the probability that a softball player gets a hit in any single at-bat is .$300$. Assuming that her chance of getting a hit at a particular time at bat is independent of her other times at bat, what is the probability that she will not get a hit until her fourth time at bat in a game?

(a) $\left(\begin{array}{l}4\\ 3\end{array}\right)(0.3{)}^{1}(0.7{)}^{3}$

(b) $\left(\begin{array}{l}4\\ 3\end{array}\right)(0.3{)}^{3}(0.7{)}^{1}$

(c) $\left(\begin{array}{c}4\\ 1\end{array}\right)(0.3{)}^{3}(0.7{)}^{1}$

(d) $(0.3{)}^{3}(0.7{)}^{1}$

(e) role="math" localid="1650371346957" $(0.3{)}^{1}(0.7{)}^{3}$

The probability that she will not get a hit until her fourth time at bat in a game is (e) $(0.3{)}^{1}(0.7{)}^{3}$.

We can use the Geometric probability equation:

$P(X=k)={q}^{k-1}p\phantom{\rule{0ex}{0ex}}=(1-p{)}^{k-1}p$

$p=\text{Probability of success}=0.300$

A geometric distribution follows the initial success among independent attempts.

We want to know how likely it is that the first hit will come on the fourth batter, thus we'll use the definition of geometric probability at $k=4$ :

localid="1650383208554" $P(X=4)=(1-0.300{)}^{4-1}(0.300)\phantom{\rule{0ex}{0ex}}=(0.700{)}^{3}(0.300)\phantom{\rule{0ex}{0ex}}=(0.7{)}^{3}(0.3)\phantom{\rule{0ex}{0ex}}=(0.7{)}^{3}(0.3{)}^{1}\phantom{\rule{0ex}{0ex}}=(0.3{)}^{1}(0.7{)}^{3}$

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