Conditional probability help?

Im trying to practice statistics for my exam but this question is bogging my mind. I have 5 cards, 10, J, Q, K and A. I need to find the probability that the first card is A, fourth is a K with the condition that third is a Q. What is really bothering me is the probability that the first is A and fourth is K. I understand that I have to make two events, A=First card is A and fourth is K. B=Third card is a Q, and then have probability of A with condition B. But I do not know how to find A and B. Especially A.

Thanks for help

2 Answers

  • 3 years ago
    Favorite Answer

    Event A: First card is Ace and fourth card is King.

    Event B: Third card is Queen.

    Definition of conditional probability:

    P(A|B) = P(A and B) / P(B)

    P(A and B) = P(1st is Ace & 3rd is Queen & 4th is King)

    P(A and B) = P(1st is Ace) * P(2nd is Jack or 10) * P(3rd is Queen) * P(4th is King)

    Remember that with each draw, the number of cards in the deck decreases by 1:

    P(A and B) = 1/5 * 2/4 * 1/3 * 1/2 = 2/120 = 1/60


    P(B) = 4/5 * 3/4 * 1/3 = 1/5


    P(A|B) = (1/60) / (1/5) = 1/12

  • nbsale
    Lv 6
    3 years ago

    You can simplify this to a different problem:

    You have 4 cards: 10 J K A, in random order.

    You pick two cards. What's the probability that you pick the Ace, then the King?

    1/4 for the Ace


    1/3 for the King

    = 1/12

    The order of the cards is immaterial. The 1 and the 4 are just random cards that could be anything. The fact that some other card is a Q just reduces the number of cards under consideration, because you know that the cards you are picking is not a Q, so they are from the other 4 cards.

    In particular, you don't need to worry about the 2nd and the 5th cards. In probability questions like this, when you have no information about some card or what's in some slot, you can usually ignore it, since those are just any random card.

    I realize this doesn't use conditional probability rules, so it might not be the best answer. All I've tried to do is to simplify the question down to its essentials.

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