2 years ago

PLZ HELP FAST

Mathematics

1 answer:

Juliette [100K]2 years ago

3 0

The answer would be 4,500 gallons left.

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(a) $P(Y'|M)\approx 0.3297$

(b) $P(Y|M')\approx 0.8323$

Step-by-step explanation:

Given table is

Yes No Don't Know Total

Men 162 92 25 279

Women 258 41 11 310

Total 420 133 36 589

According the the conditional probability, if A and B are two event then

$P(A|B)=P(\frac{A}{B})=\frac{P(A\cap B)}{P(B)}$

We need to find the following probabilities.

Let Y is the event "saying yes," and M is the event "being a man."

(a)

$P(Y'|M)=\frac{P(Y'\cap M)}{P(M)}$

$P(Y'|M)=\frac{\frac{92}{589}}{\frac{279}{589}}$

$P(Y'|M)=\frac{92}{279}$

$P(Y'|M)=0.329749103943$

$P(Y'|M)\approx 0.3297$

(b)

$P(Y|M')=\frac{P(Y\cap M')}{P(M')}$

$P(Y|M')=\frac{\frac{258}{589}}{\frac{310}{589}}$

$P(Y|M')=\frac{258}{310}$

$P(Y|M')=0.832258064516$

$P(Y|M')\approx 0.8323$

(c)

$P(Y'|M')=\frac{P(Y'\cap M')}{P(M')}$

$P(Y'|M')=\frac{\frac{41}{589}}{\frac{310}{589}}$

$P(Y'|M')=\frac{41}{310}$

$P(Y'|M')=0.132258064516$

$P(Y'|M')\approx 0.1323$

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