Answer:
E(X) = 6.0706
Step-by-step explanation:
1) Define variables
X = a random variable representing the count of heads in the first 10 tosses
Y = a random variable indicating the number of heads in tosses 4 through 10
We define the following events:
a= The event where the first coin is chosen
b= The event of selecting the second coin
c= The event of getting 2 Heads in the first two tosses
2) Relevant formulas
We need to determine E(X|c) =?
Using the total probability law helps in calculating E(Y)
E(Y) = E(Y|a) P(a|c) + E(Y|b) P(b|c)....(1)
To find P(a|c) through Bayes’ Theorem yields:
P(a|c) = P(c|a) P(a) / P(c)...(2)
Substituting P(c) via total probability gives us:
P(a|c) = [P(c|a) P(a)] /[P(c|a) P(a) + P(c|b) P(b)]... (3)
The necessary probabilities can be established:
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
Substituting values into P(a|c) gives:
P(a|c) = (0.288 x 0.5) /(0.288x 0.5 + 0.441x0.5) = 0.144/ 0.3645 = 0.39506
As P(a|c) + P(b|c) = 1, we find P(b|c) = 1 - P(a|c) = 1-0.39506 = 0.60494
Next, we calculate the expected values
E(Y|a) = 7x 0.4 = 2.8
E(Y|b) = 7x 0.7 = 4.9
Finally, substituting in equation (1) gives us
E(Y|c) = 2.8x 0.39506 + 4.9x0.60494 = 4.0706
Therefore:
E(X|c) = 2+ E(Y|c) = 2+ 4.0706 = 6.0706
approximately 0.00020