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Exercise 9.3 · Q2
Q.

From the given data, find out the probability that a randomly selected person is male, given that he owns a pet?

Have petsDo not have petsTotal
Male0.410.080.49
Female0.450.060.51
Total0.860.141
Yanam CbseNCERTSubjective· 2mImportance★★★★★est
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Reading directly off the joint-probability table, P(Male∣owns a pet)=4186≈0.477P(\text{Male}\mid\text{owns a pet})=\dfrac{41}{86}\approx 0.477.

P(M∣P)=P(M∩P)P(P)P(M\mid P)=\frac{P(M\cap P)}{P(P)}

where M=M= "person is male", P=P= "person owns a pet".

  1. Read the joint probability from the table. The cell (Male, Have pets) gives

P(M∩P)=0.41P(M\cap P)=0.41

  1. Read the marginal probability of owning a pet (column total "Have pets"):

P(P)=0.86P(P)=0.86

  1. Apply the conditional probability formula.

P(M∣P)=P(M∩P)P(P)=0.410.86P(M\mid P)=\frac{P(M\cap P)}{P(P)}=\frac{0.41}{0.86}

  1. Simplify. 0.410.86=4186≈0.4767\frac{0.41}{0.86}=\frac{41}{86}\approx 0.4767 …

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