Answer:
E(X) = 6.0706
Step-by-step explanation:
1) Define notation
X = random variable who represents the number of heads in the 10 first tosses
Y = random variable who represents the number of heads in range within toss number 4 to toss number 10
And we can define the following events
a= The first coin has been selected
b= The second coin has been selected
c= represent that we have 2 Heads within the first two tosses
2) Formulas to apply
We need to find E(X|c) = ?
If we use the total law of probability we can find E(Y)
E(Y) = E(Y|a) P(a|c) + E(Y|b)P(b|c) ....(1)
Finding P(a|c) and using the Bayes rule we have:
P(a|c) = P(c|a) P(a) / P(c) ...(2)
Replacing P(c) using the total law of probability:
P(a|c) = [P(c|a) P(a)] /[P(c|a) P(a) + P(c|b) P(b)] ... (3)
We can find the probabilities required
P(a) = P(b) = 0.5
P(c|a) = (3C2) (0.4^2) (0.6) = 0.288
P(c|b) = (3C2)(0.7^2) (0.3) = 0.441
Replacing the values into P(a|c) we got
P(a|c) = (0.288 x 0.5) /(0.288x 0.5 + 0.441x0.5) = 0.144/ 0.3645 = 0.39506
Since P(a|c) + P(b|c) = 1. With this we can find P(b|c) = 1 - P(a|c) = 1-0.39506 = 0.60494
After this we can find the expected values
E(Y|a) = 7x 0.4 = 2.8
E(Y|b) = 7x 0.7 = 4.9
Finally replacing the values into equation (1) we got
E(Y|c) = 2.8x 0.39506 + 4.9x0.60494 = 4.0706
And finally :
E(X|c) = 2+ E(Y|c) = 2+ 4.0706 = 6.0706
