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Consider a large village, where only two newspapers $P_1$ and $P_2$ are available to the families. It is known that the proportion of families

  1. not taking $P_1$ is $0.48$,
  2. not taking $P_2$ is $0.58$, 
  3. taking only $P_2$ is $0.30$.

The probability that a randomly chosen family from the village takes only $P_1$ is

  1. $0.24$
  2. $0.28$
  3. $0.40$
  4. can not be determined

2 Answers

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Taking P1 = 0.52

Taking P2 = 0.42

P1 ⋂ P2 = P2 - Only P2

P1 ⋂ P2 = 0.42 - 0.30 = 0.12

Only P1 = P1 - (P1 ⋂ P2)

Only P1 = 0.52 - 0.12 = 0.40
1 votes
1 votes

Answer: 0.40

Only P2=0.30

Not P1=Only P2  +  $\overline{P1 \bigcup P2}$=0.48

$\overline{P1 \bigcup P2}$=0.48-0.3=0.18

Not P2=Only P1  +  $\overline{P1 \bigcup P2}$=0.58

Only P1=0.58-0.18=0.40

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