There are two misshapen coins in a box; their probabilities for landing on heads when they are flipped are, respectively,.4 and. 7. One of the coins is to be randomly chosen and flipped 10 times. Given that two of the first three flips landed on heads, what is the conditional expected number of heads in the 10 flips?

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