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Darkknighta26

Darkknighta26

04.03.2021 •

Geography

Japan ended its practice of isolationism in what year? A. 1945
B. 1853
C. 1610
D. 1787

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Ответ:

frankcaroccio0203

04.03.2021

d

Explanation:

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Ответ:

swaise3300

04.03.2021

B

Explanation:

teachers mention it whenever the topic is brought up

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Ответ:

ryliepeloquinf

25.06.2022

Check the explanation

Step-by-step explanation:

The given problem can be modeled into a geometric distribution as follows.

Suppose, X be the number of at-bats required to get a home run. As each at-bat is independent.

$\therefore X \sim Geom (p=0.16)$

$\therefore P(X=x)=(1-p)^{x-1}p ;x=1,2,3,.....$

A.

Required probability is given by

P(X=5)=(1-0.16)^{5-1}0.16=0.07965942

B.

We have to calculate E(X).

$E(X)=\sum_{x}x*P(X=x)=1p+2(1-p)p+3(1-p)^2p+4(1-p)^3p+...$

$(1-p)E(X)=\sum_{x}x*P(X=x)=1(1-p)p+2(1-p)^2p+3(1-p)^3p+...$

Subtracting later from earlier we get,

$\left \{ 1- \(1-p) \right \}E(X)=p+(1-p)p+(1-p)^2p+(1-p)^3p+...$

$\Rightarrow pE(X)=p\left \{1+(1-p)+(1-p)^2+(1-p)^3+... \right \}$

$\Rightarrow E(X)=\frac{1}{1-(1-p)}=\frac{1}{p}$

So, expected number of at-bats until home run = 1/0.16 = 6.25

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