Conditiona expectatio : Tosses of a biased coin A box contains 2 biased coins, one with probability of 0.4 of head and other with probability 0.7 of head. A coim in randomly chose and tossed 10 times. Evaluate the conditional expectation of the number of heads given that the first 2 of the first 3 tosses were head.
My attempt:
Let $X$ denote the number of heads that appeared on those tosses. Define
$$
X_i=1 \text{if a head appears}$$
$$X_i=0, otherwise$$
If a coin is randomly chosen, the probability of head should be
$$\frac{1}{2} 0.4 + \frac{1}{2} 0.7$$
Then 
$$P(X_i=1)=0.55$$
Now, we can see that 
$$X=X_1+...+X_{10}$$
Note that there are 4 cases where 2 heads (H) appears on the first 3 tosses:
$$HHH,HHT,HTH,THH$$
If $ Y  $is the event where 2 heads appear on the first 3 tosses, then, for example, 
$$E(X_1|Y)=P(HHH)+P(HHT)+P(HTH) = (0.55)^3 + 2(0.55)^20.45 $$
We should have the same for $ X_2, X_3.$ But,$ X_4, ..., X_{10}$ are independent of Y (considering independent tosses), then
$$E[X_i|Y]=E[X_i]=0.55, i=4,...,10$$
Now, we shpuld sum and obtain the conditional expectation of $X.$ But I am getting 5.16, while the answer is 6.06. What is wrong?
Thanks!
 A: The question is a little ambiguous.  When you say "the first 2 of the first 3 tosses are heads," that is clearly a typo; based on the answer you gave, I assume it should read "exactly 2 of the first 3 tosses are heads."  (If it should instead read "at least 2 of the first 3 tosses are heads," the answer will differ.)
Your main error is treating the tosses as independent of each other: All the tosses depend on which coin you chose, and are independent given that choice, which isn't the same thing.  So
$$
P(2H \;|\; A)=3(0.4)^2(0.6) = 0.288, \\
P(2H \;|\; B)=3(0.7)^2(0.3) = 0.441,
$$
where $A$ means you chose the tail-weighted coin and $B$ means you chose the head-weighted coin.  In case $A$, the expected number of heads is clearly $2 + 0.4\times 7 = 4.8$; and in case $B$ it's $2 + 0.7\times 7=6.9$.  So you can write
$$
E[N_H\;|\;2H]=E[N_H\;|\;A]P(A\;|\;2H)+E[N_H\;|\;B]P(B\;|\;2H)\\ =4.8P(A \; | \; 2H)+6.9(1-P(A\;|\;2H))=6.9-2.1P(A\;|\;2H).
$$
Moreover,
$$
P(A\;|\;2H)=\frac{P(2H\;|\;A)P(A)}{P(2H)}=\frac{0.288 \cdot0.5}{0.288\cdot 0.5 + 0.441 \cdot 0.5}=\frac{32}{81}.
$$
The result is then
$$
E[N_H \;|\; 2H] = 6.9 - 2.1\times\frac{32}{81}=6.07.
$$
A: The fact that you had more heads than tails in the first three tosses tells you that the coin with probability $0.7$ was more likely chosen. This affects the conditional expectation of all the tosses, not just the first three.
Judging from the result you were supposed to get, I think the "two of the first three" condition is supposed to mean that there were exactly two heads in the first three tosses.
The prior probability of this is $p_1 = 3(0.7)^2(0.3)$ for the $0.7$ coin but 
$p_2 = 3(0.4)^2(0.6)$ for the $0.4$ coin.
This makes the probability $\frac{p_1}{p_1 + p_2} = \frac{49}{81}$ 
that the $0.7$ coin is the one that was chosen. 
The expected number of heads is therefore
$$2 + 7 \left(0.7 \frac{49}{81} + 0.4 \frac{32}{81} \right).$$
(I get approximately $6.0704$ when computing this value; I'm not sure why the answer key says $6.06.$ But other interpretations of the "two of three" condition seem to take us farther away from that value.)
A: As the question is written, you are saying that HH was gotten for the first $2$ coin tosses.  I would say from only $2$ coin tosses that no conclusion can be drawn as to which biased coin was chosen because either one could have "easily" generated HH.
Thus my answer would be $2 + 8 * 0.55 = 6.4$
