I have this statement:
In a bag there are four cards marked with the letter $A$ and six cards marked with the letter $B$, all of equal shape and size. A game consists of taking out two random cards, one by one and with replacement, where if both correspond to type $A$, then you win $\$1,000$; if both cards are different, you earn $\$200$; and if both cards have the letter $B$, then $\$1,500$ is lost. If you want to participate in the game, then it is estimated, from the calculation of the expected value, that the result of the game will be:
A) Lose $\$256$
B) Lose $\$284$
C) Win $\$100$
D) Win $\$256$
E) Neither win nor lose
My development was:
First i need to get the probability of get $AA$ or $AB$ or $BA$ or $BB$
The cases for $AA$ is equal to $\binom{4}{2}_{rep} = 10$
The cases for $AB$ is equal to $\binom{4}{1}\cdot\binom{6}{1}=24$
The cases for $BA$ is equal to $\binom{6}{1}\cdot\binom{4}{1}=24$
The cases for $BB$ is equal to $\binom{6}{2}_{rep} = 21$
The total cases of choose two cards of a total of $10$ is equal to $10 + 24 + 24 + 21 = 79$
Thus, $1000(\frac{10}{79}) + 200(\frac{48}{79}) - 1500(\frac{21}{79}) = E(X)$
$E(X) = \approx-149$ or Lose $\$149$, but this result don't appear in the answers. So, what ins wrong with my development? Thanks in advance.
A strong suggestion that your numbers aren't right: you say there are $79$ possible ways you could experience this game. That's a very prime number in a context where I would expect a highly composite number.
I advise instead discarding this case-counting thing and just working with probabilities directly (since you're drawing cards with replacement). The probability of $A$ is $\frac{2}{5}$, so the probability of $AA$ is $\left(\frac{2}{5}\right)^2$.