Show work please A probability space S has five elements- call them a, b, c, d a

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Show work please

A probability space S has five elements- call them a, b, c, d and e. Their respective probabilities are .1, .1, .2, .2 and .4. The random variable X takes on the values 4, 6, 5, 10 and -10 on these five elements, respectively. Compute the expected value of X and the variance of X. If X represents the amount you win or lose in one play of a game, and X(n) is the result of playing this game n times, independently, compute the probability that X(2) > 0 and the probability that X(3) > 0. Compute the expected value of X(10000) and the Variance of X(10000). Find the probability that X(10000 )> 0; that it is >800; that it is > 1000, and that it is < 2000.

Explanation / Answer

Ans:

Mean or expectation=x.p(x)=4*0.1+6*0.1+5*0.2+10*0.2-10*0.4=0.4+0.6+1+2-4=0

Var(x)=(0-4)^2*0.1+(0-6)^2*0.1+(0-5)^2*0.2+(0-10)^2*0.2+(0+10)^2*0.4=70.2

Now ,P(X(2)>0)

it is possible when, we dont get -10 in first two plays,as if in one of the two plays we get -10 it will give us X(2)<0.

Hence,not getting -10 is with 4/5 chance

So ,P(X(2)>0)=0.6*0.6=0.36

P(X(3)>0)=2/3*0.6+1/3*0.4=0.532

Now E[X(10000)]=10000*E(X)=0

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