Question

In a newspaper poll concerning violence on television, 589 people were asked, "What is your opinion of the amount of violence on prime-time television — is there too much violence on television?"
Yes No Don't Know Total
Men 162 92 25 279
Women 258 41 11 310
Total 420 133 36 589
Use the data in the table above to find the following probabilities, where Y is the event "saying yes," and M is the event "being a man." (Round your answers to four decimal places.)

(a) p(Y ' | M)


(b) p(Y | M')


(c) p(Y ' | M')

Answer 1

Answer:

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

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

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

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)

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

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

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

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

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