Suppose we toss a biased coin independently until a random time N independent of the outcomes of the tosses. Where N takes values 1,2,3 with probability 0.3, 0.5, 0.2. Find E(X1 + · · · XN) where Xi = 1 if head on ith toss with probability 0.55 and zero otherwise, (for i = 1, · · · , N).

E(X1+...+XN) = 1.045

Step-by-step explanation:

For a fixed number k, the mean of a random variable that counts the amount of success in k tries each one with success 0.55 is k*0.55 (this random variable is binomial).

Note that Xi is 1 only when the toss is heads and 0 otherwise, so with X1+...+Xn we are just counting the amount of tosses (success). Now, lets compute the expected value using conditionals:

E(X1++XN) = E(X1+...+Xn | N = 1) * P(N=1) + E(X1+...+Xn | N = 2) * P(N=2) + E(X1+...+Xn | N = 3) * P(N=3) = 0.55*1 * 0.3 + 0.55*2 * 0.5 + 0.55*3 * 0.2 = 1.045  

step-by-step explanation:

compare

on monday rodney ate 1/4th of the sandwich = 3/12

on tuesday he ate 1/3rd of the sandwich = 4/12

therefore in total he has eaten = 3/12 + 4/12 = 7/12

thus we can say that he has eaten 7/12 of the sandwich on monday and tuesday alltogether.

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