Question

Two newspaper X & Y are published in a certain city . It is estimated from a survey that 16% read X, 14% read Y & 5% read both. Find the probability that a randomly selected person i) does not read any newspaper ii) reads only y.​

Answer 1

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Answer 2

Given :

Two newspaper X & Y ,

Probability that people read newspaper X = P(X) = $\frac{16}{100} = \frac{4}{25}$

Probability that people read newspaper Y = P(Y) = $\frac{14}{100} = \frac{7}{50}$

Probability that people read newspaper X and Y = P(X∩Y) = $\frac{5}{100} = \frac{1}{20}$

To Find :

(i)Probability that a person does not read any newspaper = P'(XUY)

(ii)Probability that person reads only Y = P(Y) - P(X∩Y)

Step-by-step explanation:

(i)  P'(XUY) = 1 - P(XUY) = 1 - (P(X) +P(Y) - P(X∩Y)) = $1 - (\frac{4}{25} + \frac{7}{50} -\frac{1}{20} )$

    P'(XUY) = 0.25

(ii)  P(Y) - P(X∩Y) = $\frac{7}{50} - \frac{1}{20}$ = 0.09

Hence ,

Probability that a person does not read any newspaper = P'(XUY)  = 0.25

Probability that person reads only Y = P(Y) - P(X∩Y) = 0.09

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