0 votes 0 votes 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 not taking $P_1$ is $0.48$, not taking $P_2$ is $0.58$, taking only $P_2$ is $0.30$. The probability that a randomly chosen family from the village takes only $P_1$ is $0.24$ $0.28$ $0.40$ can not be determined Probability isi2018-mma engineering-mathematics probability + – akash.dinkar12 asked May 11, 2019 akash.dinkar12 1.3k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
2 votes 2 votes 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 kraken_wizard answered Mar 1, 2020 kraken_wizard comment Share Follow See all 0 reply Please log in or register to add a comment.
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 sakharam answered May 11, 2019 sakharam comment Share Follow See all 0 reply Please log in or register to add a comment.