In a certain region, 46% of the population is female.It is known that 5% of males and 2% of females are left-handed.A person is chosen at random and found to be left-handed.What is the probability that this person is a male?

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Answer 1 · 2018-07-15T10:04:27

#"The Reqd. Prob."=135/181~~74.59%#.

Let, #F, M, and, L# be the events that a randomly chosen person

from the region is a Female, Male, and, Left handed, resp.

From what is given, we have,

#P(F)=46/100, P(L/M)=5/100, and P(L/F)=2/100#.

We deduce #P(M)=P(F')=1-P(F)=1-46/100=54/100#.

Now, #P(L)=P(F)P(L/F)+P(M)P(L/M)#,

#=46/100*2/100+54/100*5/100#,

#=(92+270)/10000#.

# rArr P(L)=362/10000#.

By Definition,

#P(M/L)=(P(MnnL))/(P(L)), &, P(L/M)=(P(L nnM))/(P(M))#.

#:. P(M/L)-:P(L/M)=(P(M))/(P(L))#.

#rArr"The Reqd. Prob."=P(M/L)=P(L/M)*(P(M))/(P(L))#,

#={(5/100)(54/100)}/(362/10000)#,

#=135/181#.

#:. "The Reqd. Prob."=135/181~~74.59%#.