Ptrobability (poisson theorem) help?

1. Special laboratory equipments of a certain diameter are kept in the laboratory store room.A survey reveals that they are drawn from the store, an average, 1.5 times a day.Assuming the decimal follows a poisson distribution,determine for a period of 200 working days.


a.)The number of days when none of the equipment are used.


b.)the probability that four of the equipments are used


c.)the probability that at most 4 of the equipments are used


d.)the number of days when at least 3 of the equipments are used


e.)the number of days when no more than 3 of the equipments are used.





2. In a gambling game,a man is paid $215 if he gets all the heads or all tails when three coins are tosssed and he will pay out $129 if either one or two heads show.what is his expected gain?





please help me,i really dont know the right answer and solution


please.....thank you

Ptrobability (poisson theorem) help?
If they are drawn at an average of 1.5 times a day, the average is 200*1.5, or 300 times in 200 days. On any day, the use of those equipments follows a Poisson distribution with lambda = 1.5


Then the probability that K of the equipment are used on any day is given by:


P(K) = 1.5^K*e^(-1.5)/K!





a). On any single day, the probability that none of the equipment are used will be P(0) = 1.5^0*e^(-1.5)/0! = e^(-1.5) = 22.3%


Over a 200 days period, you will have an average of 0.223*200 = 44.6 days when none of the equipment are used. Round up to 45 days.





b) P(4) = 1.5^4*e^(-1.5)/4! = 4.7% on any day.





c) P(at most 4) = P(0) + P(1) + P(2) + P(3) + P(4)


P(0) = 1.5^0*e^(-1.5)/0!


P(1) = 1.5^1*e^(-1.5)/1!


P(2) = 1.5^2*e^(-1.5)/2!


P(3) = 1.5^3*e^(-1.5)/3!


P(4) = 1.5^4*e^(-1.5)/4!


P(at most 4)= e^(-1.5)*(1/1 + 1.5/1 + 1.5^2/2 + 1.5^3/6 + 1.5^4/24)


P(at most 4)= e^(-1.5)*4.398 = 98.1%





d) P(at least 3) = 1 - P(at most 2)


P(at most 2) = P(0) + P(1) + P(2) = e^(-1.5)(1/1+ 1.5/1 + 1.5^2/2)





P(at least 3) = 1 -e^(-1.5)*3.625 = 19.1%





Over 200 days, you will have an average of .191*200 days when at least 3 of the equipment are used, or 38 such days (rounded down).





e) P(at most 3) = P(0) + P(1) + P(2) + P(3) = e^(-1.5)(1/1+ 1.5/1 + 1.5^2/2 + 1.5^3/6) = e^(-1.5)*4.1875 = 93.4%.


Over 200 days, you will have an average of .934*200 days when at most 3 of the equipment are used, or 187 such days (rounded up).





2. on the toss of three coins, the odds are:


3 tails: 1/2*1/2*1/2 = 1/8


3 heads: 1/2*1/2*1/2 = 1/8


2 tails, one head: 1/2*1/2*1/2*3 = 3/8 (three possibilities of picking one head amongst three coins)


2 heads, one tail: 1/2*1/2*1/2*3 = 3/8


His winning odds are then 1/8+1/8 = 1/4 (three heads or three tails)


His losing odds are 3/6+3/8 = 3/4





On average he will earn $215 once every 4 tries and lose $129 the rest of the time.


His expected gain is P(win)*Gain(win) + P(lose)*Gain(lose) = 1/4*215 + 3/4*(-129) = -$43


His expected gain is a net loss of $43.

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