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!
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
$$
E[N_H \;|\; 2H] = 6.9 - 2.1\times\frac{32}{81}=6.07.
$$
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 $2 + 8 * 0.55 = 6.4$