Question #210301

A bag contains so many pink and yellow balls. A person can have 2 balls. Given that one of the balls is pink, what

is the probability that the other one is also pink?

(b) One hundred cars enter for a road-worthiness test which is in 2 parts mechanical and electrical. A car can only

pass if it passes both parts. Half the cars fail the electrical test and 62 pass the mechanical. 15 pass the electrical

and fail the mechanical test.

Find the probability that:

(i) A car chosen at random given that it has failed, failed the electrical test only

(ii) The Probability of fails on one test only will be


Expert's answer

a) Let n=n= the number of pink balls in bag. Then there are nn yellow balls in a bag.


P(II pinkI pink)=n12n1P(II\ pink|I\ pink)=\dfrac{n-1}{2n-1}

b) Draw the Venn diagram


N(M)=62,N(E)=50,N(EM)=15N(M)=62, N(E)=50, N(E\cap M')=15

N(ME)=N(E)N(EM)N(M\cap E)=N(E)-N(E\cap M')

=5015=35=50-15=35

N(ME)=N(M)N(ME)N(M\cap E')=N(M)-N(M\cap E)

=6235=27=62-35=27


N(ME)=N(M)+N(E)N(ME)N(M\cup E)=N(M)+N(E)-N(M\cap E)


=62+5035=77=62+50-35=77


N(ME)=100P(ME)N(M'\cap E')=100-P(M\cup E)

=10077=23=100-77=23

(i)


P(EMME)=1577P(E\cap M'|M\cup E)=\dfrac{15}{77}

(ii)


P(ME or EM)=27+15100=0.42P(M\cap E'\ or\ E\cap M')=\dfrac{27+15}{100}=0.42


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