For example, in a standard 52-card deck with two colours it is impossible to draw a card that is both red and a club because clubs are always black. If just one card is drawn from the deck, either a red card (heart or diamond) or a black card (club or spade) will be drawn. When A and B are mutually exclusive, P(A ∪ B) = P(A) + P(B) .Stats: Probability Rules. To find the probability of drawing a red card or a club, for example, add together the probability of drawing a red card and the probability of drawing a club. In a standard 52-card deck, there are twenty-six red cards and thirteen clubs: 26/52 + 13/52 = 39/52 or 3/4. One would have to draw at least two cards in order to draw both a red card and a club. The probability of doing so in two draws depends on whether the first card drawn were replaced before the second drawing, since without replacement there is one fewer card after the first card was drawn. The probabilities of the individual events (red, and club) are multiplied rather than added. The probability of drawing a red and a club in two drawings without replacement is then 26/52 × 13/51 × 2 = 676/2652 , or 13/51. With replacement, the probability would be 26/52 × 13/52 × 2 = 676/2704 , or 13/52.
For example, in a standard 52-card deck with two colours it is impossible to draw a card that is both red and a club because clubs are always black. If just one card is drawn from the deck, either a red card (heart or diamond) or a black card (club or spade) will be drawn. When A and B are mutually exclusive, .Stats: Probability Rules. To find the probability of drawing a red card or a club, for example, add together the probability of drawing a red card and the probability of drawing a club. In a standard 52-card deck, there are twenty-six red cards and thirteen clubs: 26/52 + 13/52 = 39/52 or 3/4. One would have to draw at least two cards in order to draw both a red card and a club. The probability of doing so in two draws depends on whether the first card drawn were replaced before the second drawing, since without replacement there is one fewer card after the first card was drawn. The probabilities of the individual events (red, and club) are multiplied rather than added. The probability of drawing a red and a club in two drawings without replacement is then , or 13/51. With replacement, the probability would be , or 13/52.
D
P(A ∪ B) = P(A) + P(B)
D
26/52 × 13/51 × 2 = 676/2652
D
26/52 × 13/52 × 2 = 676/2704