Mathematics

Question

In a standard deck of 52 playing cards, 13 cards are clubs, and 3 of the clubs are "face" cards (K, Q, J). What is the probability of drawing one card that is:


A) A club or a face card? Is this event a union or an intersection?


B) A club and a face card? Is this event a union or an intersection?


C) Not a club and not a face card?

1 Answer

  • A) This event is a union. If a card is of the suit of clubs or a face card or both, then it belongs to the set of "cards with clubs or a face".

    For the probability in question, we use the inclusion/exclusion principle:

    P(club or face) = P(club) + P(face) - P(club and face)

    That is, we add the probabilities of drawing a club and of drawing a face card, and subtract the probability of both conditions being met because, in a certain sense, we've double-counted the event of the card being both of clubs and a face. The sets "club" and "face" are not disjoint; there is overlap between them.

    Then we have

    P(club) = 13/52

    P(face) = 12/52

    P(club and face) = 3/52

    so that

    P(club or face) = 13/52 + 12/52 - 3/52 = 22/52 = 11/26

    B) This event is an intersection. If a card is of the suit of clubs and a face card, then it belongs to both the set of clubs and the set of face cards.

    We already found the probability of the intersection above:

    P(club and face) = 3/52

    C) If a given card is not a club and not a face card, this is the same as saying the card is neither a club nor a face card. In other words,

    P(not club and not face) = P(not(club or face))

    and this probability is complementary to the probability that a card is either a club or face, which is to say

    P(not(club or face)) = 1 - P(club or face)

    Using the result from part (A), we have

    P(not club and not face) = 1 - 11/26 = (26 - 11)/26 = 15/26

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