Find the conditional probability that the deck was fixed given that the inspector lost all 3 games. An inspector has been informed that a certain gambling casino uses a "fixed" deck of cards onequarter of the time in its blackjack games. With a fair deck, the probability of the casino winning a particular hand of blackjack is .52, but with a fixed deck the probability of the casino winning a particular hand is .75 . The inspector visits the casino and plays 3 games of blackjack (from the same deck of cards), losing all of them. Find the conditional probability that the deck was fixed given that the inspector lost all 3 games.
I almost did the entire question right except for one small mistake which I can't clear out. Why is the probability of him losing all 3 games :
$0.75^3 *0.25 + 0.52^3*0.75 $ instead of $(0.75*0.25)^3 + (0.52*0.75)^3$. The latter is what I chose and think makes more sense in my head. Can anyone clear this confusion out please? If i get this step correct my answer comes out right.
 A: Consider first that there are two cases: either the deck is fixed, or it isn't. It is with probability $0.25$, and it isn't with probability $0.75$. 
Then you work each case in turn. If the deck is fixed, the probability of the inspector losing all three is $0.75^3$, and so in this case we have $0.25 \cdot 0.75^3$. 
If the deck isn't fixed, then the probability of the inspector losing all three is $0.52^3$, and so in this case we have $0.75 \cdot 0.52^3$. 
Now just add for total probability of losing.
A: Apply what some textbooks call the Law of Total Probability!
Let $ A $ be the event that the inspector loses all three games. Let us also assume that the games are independent.
In order to evaluate this probability, we can condition on the deck that the casino uses for these three games. So we have the following equation by the Law of Total Probability.
$$
P(A) = \mathbb{E}(P(A \mid \text{deck}))
$$
And now, to be totally explicit, we say that
$$
\text{deck} = \begin{cases}
\text{fixed} & \text{w.p. } 0.25 \\
\text{fair} & \text{w.p. } 0.75
\end{cases}
$$
Do you see how to continue? $ P(A \mid \text{deck} = \text{fixed}) $ is the probability that the inspector loses
all three games given the deck is fixed. $ P(A \mid \text{deck = fair}) $ is the probability
that the inspector loses all three games given that the deck is fair. By
the definition of expectation, we have
$$
\mathbb{E}(P(A \mid \text{deck})) = P(\text{deck = fair})P(A \mid \text{deck = fair}) + P(\text{deck = fixed})P(A \mid \text{deck = fixed})
$$
which is the answer you desire.
