# Maths Question

1. 1. a) Suppose that avoter poll is taken in three states. In state A, 50% of voters support the liberal candidate; in state B, 60% of the voters support the liberal candidate; and in state C, 35% of the voters support theliberal candidate. Of the totalpopulation of the three states, 40% live in state A, 25% live in state B and 35% live in state C. Given that a voter supports the liberal candidate, whatis the probability that he/she lives in state B? b) If it rains, an umbrella salesman canearn \$300 per day. If it is cloudy, he can earn \$25 perday. If it is fair, he can lose \$60 per day. What is the salesman’s expectation if the probability of rain is 0.3and fair is 0.5 ? c) Bychecking several cartons containing large numbers of eggs, it was found thatthe average number of broken eggs a carton was 2.Find: (i) theprobabilities of finding a carton containing 0, 1 and 2 brokeneggs; and (ii) the probability of finding a cartoncontaining 3 or more broken eggs. Thanks!

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1) Given that a voter supports the liberal candidate, whatis the probability that he/she lives in state B?

Probability that a voter supports the liberal candidate:

50%*40%+60%*25%+35%*35%

=0.2+0.15+0.1225

=0.4725

P(the voter lives in State B & supports liberal candidate ∣the voter supports the liberal candidte)

= 60%*25% /0.4725

=0.15/0.4725

=0.317 (corr. to 3 sig. fig.) //

b) If it rains, an umbrella salesman canearn \$300 per day. If it is cloudy, he can earn \$25 perday. If it is fair, he can lose \$60 per day. What is the salesman’s expectation if the probability of rain is 0.3and fair is 0.5 ?

Salesman'ts expectation =\$300*0.3+ \$25*0.2 - \$60*0.5

=\$90+\$5 -\$30

=\$65 //

c) Bychecking several cartons containing large numbers of eggs, it was found thatthe average number of broken eggs a carton was 2.

Find: (i) theprobabilities of finding a carton containing 0, 1 and 2 brokeneggs; and

Using the poisson distribution

P(a carton containing 0)= e^(-2) (2)^0/0! =0.1353//

P(a carton containing 1)=e^(-2) (2)^1/1!=0.2707//

P(a carton containing 2)=e^(-2) (2)^2/2!=0.2707//

(ii) the probability of finding a cartoncontaining 3 or more broken eggs.

P(a carton containing 3 or more broken eggs)

=1-0.1353-0.2707-0.2707

=0.3233//

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