From a standard deck, what are the probabilities of A. picking a red card, B. picking a spade, C. picking an Ace?

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From a standard deck, what are the probabilities of A. picking a red card, B. picking a spade, C. picking an Ace?

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Answer 1 · 2017-04-11T09:05:18

$P (A) = \frac{26}{52} = \frac{1}{2}$ $P(A)=26/52=1/2$

$P (B) = \frac{13}{52} = \frac{1}{4}$ $P(B)=13/52=1/4$

$P (C) = \frac{4}{52} = \frac{1}{13}$ $P(C)=4/52=1/13$

Explanation:

When calculating probabilities, we're looking at a ratio:

$\frac{number of ways to meet the conditions}{number of ways we can pick}$ $"number of ways to meet the conditions"/"number of ways we can pick"$

For the three questions, the number of ways we can pick a card is 52.

For $P (A) = \frac{red card}{52}$ $P(A)="red card"/52$ , how many red cards are there? In a standard deck, 26 cards are red and the other half are black, and so:

$P (A) = \frac{26}{52} = \frac{1}{2}$ $P(A)=26/52=1/2$

For $P (B) = \frac{spade}{52}$ $P(B)="spade"/52$ , how many spades are there? In a standard deck, 13 cards are spades (with 3 other sets of 13 cards being clubs, hearts, and diamonds), and so:

$P (B) = \frac{13}{52} = \frac{1}{4}$ $P(B)=13/52=1/4$

For $P (C) = \frac{ace}{52}$ $P(C)="ace"/52$ , how many aces are there? In a standard deck, 4 cards are aces (with the other ordinals being 2 through 10, Jack, Queen, King) with each of the ordinals having a card in each of the 4 suits):

$P (C) = \frac{4}{52} = \frac{1}{13}$ $P(C)=4/52=1/13$

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From a standard deck, what are the probabilities of A. picking a red card, B. picking a spade, C. picking an Ace?

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