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Q. 1

Expert-verified

The Practice of Statistics for AP

Found in: Page 667

Book edition 4th

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

Pages 809 pages

ISBN 9781319113339

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Short Answer

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) $(\begin{array}{l} 4 \\ 3 \end{array}) (0.3)^{1} (0.7)^{3}$
(b) $(\begin{array}{l} 4 \\ 3 \end{array}) (0.3)^{3} (0.7)^{1}$
(c) $(\begin{matrix} 4 \\ 1 \end{matrix}) (0.3)^{3} (0.7)^{1}$
(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) $(0.3)^{1} (0.7)^{3}$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given Information

We can use the Geometric probability equation:

$P (X = k) = q^{k - 1} p = (1 - p)^{k - 1} p$

Step 2: Explanation

$p = Probability of success = 0.300$

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" $P (X = 4) = (1 - 0.300)^{4 - 1} (0.300) = (0.700)^{3} (0.300) = (0.7)^{3} (0.3) = (0.7)^{3} (0.3)^{1} = (0.3)^{1} (0.7)^{3}$

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