### Instruction: Two cards are chosen at random from a standard 52-card deck.  What is the probability that the first card is a heart and the second card is a 10?### Response: This is a problem about dependent events, meaning that the outcome of the first event affects the outcome of the second event.
To find the probability of two dependent events, I can use the rule P(A and B) = P(A) * P(B | A), where P(B | A) means the probability of B given A.
In this case, A is the event that the first card is a heart, and B is the event that the second card is a 10.
So I need to find P(A) and P(B | A).
P(A) is easy, it is just the fraction of hearts in the deck, which is 13/52 or 1/4.
P(B | A) is trickier, because it depends on whether the first heart was a 10 or not.
If the first heart was a 10, then there are only 3 10s left in the deck, out of 51 cards.
So P(B | A) = 3/51 in that case.
If the first heart was not a 10, then there are still 4 10s left in the deck, out of 51 cards.
So P(B | A) = 4/51 in that case.
To account for both cases, I need to use the law of total probability, which says that P(B | A) = P(B | A and C) * P(C | A) + P(B | A and not C) * P(not C | A), where C is some other event.
In this case, C is the event that the first heart was a 10.
So P(B | A) = P(B | A and C) * P(C | A) + P(B | A and not C) * P(not C | A).
I already know P(B | A and C) and P(B | A and not C), they are 3/51 and 4/51 respectively.
P(C | A) is the probability that the first heart was a 10, given that it was a heart.
That is just 1/13, since there is only one 10 of hearts in the deck.
P(not C | A) is the probability that the first heart was not a 10, given that it was a heart.
That is just 12/13, since there are 12 other hearts in the deck.
So P(B | A) = (3/51) * (1/13) + (4/51) * (12/13).
Simplifying, I get P(B | A) = (3/663) + (48/663) = (51/663) = (17/221).
Now I have P(A) and P(B | A), so I can use the rule P(A and B) = P(A) * P(B | A).
P(A and B) = (1/4) * (17/221) = (17/884).