What is the probability of selecting an ace or a spade from a deck of 52 cards?

Probability gives the chances of how likely an event is to occur.

Answer: The probability of drawing a card from a standard deck and choosing a king or an ace is (1/13) × (4/51)

Let us go through the explanation to understand it better.

Explanation:

Total number of cards in a standard deck = 52

Number of Aces in a deck= 4

Number of Kings in a deck = 4

Probability of drawing a card from a standard deck and choosing a king or an ace = Probability of getting an Ace + Probability of getting a King

Probability of drawing an Ace at random = 4/52 = 1/13

Now, the probability of drawing a King at random = 4/52 = 1/13

hence, the required probability = 1/13 + 1/13 = 2/13

Therefore, the probability of drawing a card from a standard deck and choosing a king or an ace is 2/13.

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I'm trying to study for my statistics final by looking over the exams we've taken this semester and I can't quite figure out what I did wrong here. The question is: Two cards are randomly drawn from a well-shuffled standard deck a. What is the probability of getting an Ace and a Spade (in any order) when the cards are drawn WITH REPLACEMENT? My answer was $\frac{4}{52} * \frac{13}{52} $ I got a point taken off for this for some reason (2 point question). I don't really understand why? The question is with replacement, so the probability of drawing an ace would be 4/52 and the probability of drawing a spade would be 13/52. What am I missing?

asked Dec 9, 2019 at 23:56

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The question looks fuzzy, but my take is as follows: Three cases.

$P_1=\frac{1}{52}\times\frac{16}{52}$ - first card is ace of spades.

$P_2=\frac{3}{52}\times\frac{13}{52}$ - first card is other ace.

$P_3=\frac{12}{52}\times\frac{4}{52}$ - first card is other spade.

Total $P_1+P_2+P_3=\frac{103}{2704}$

answered Dec 10, 2019 at 1:22

herb steinbergherb steinberg

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Solution

The correct option is BFalseProbability of an event E, P(E) =number of favourable outcomestotal number of outcomes Since there are four aces in a deck of 52 cards, the number of favourable outcomes of getting an ace is 4. Hence the probability of drawing an ace is = 452 = 113. Therefore the given statement is false.

Answer

Verified

Hint: In a deck of cards there are 13 spades and 4 aces. Then we have to find the probability of drawing spades from deck of cards and probability of drawing aces from deck of cards, now there is a probability of getting aces of spades so remove that probability from addition of those two probabilities. Probability of drawing an ace or a spade or both from a deck of card is denoted by \[P(A\cup B)\]Complete step-by-step answer:
We know that there are 52 cards in total.
Hence, the total number of possible outcomes is 52
 Probability of drawing an ace or a spade or both from a deck of card is
The total number of spades in a deck of cards is 13
The probability of drawing spades in a deck of cards is \[P(A)=\dfrac{13}{52}\]
The total number of aces cards in a deck of cards is 4
The probability of drawing aces in a deck of cards is \[P(B)=\dfrac{4}{52}\]
The total number of aces of spade is 1
\[P(A\cap B)=\dfrac{1}{52}\]
Probability of drawing an ace or a spade or both from a deck of card is
\[P(A\cup B)=P(A)+P(B)-P(A\cap B)\]
= \[\dfrac{13}{52}+\dfrac{4}{52}-\dfrac{1}{52}=\dfrac{16}{52}=\dfrac{4}{13}\]
Probability of drawing an ace or a spade or both from a deck of cards is \[\dfrac{4}{13}\].

Note: From the venn diagram we can obtain the relation that is \[P(A\cup B)=P(A)+P(B)-P(A\cap B)\]. Here \[P(A\cup B)\]represents probability of happening of events A and B or probability of happening both.By using the formula we are able to solve the problem. We know that the probability is the ratio of total number of desired outcomes to the total number of all possible outcomes.

What is the probability of pulling a spade or an ace from the deck?

How many Ace of Spades are in a deck of 52 cards?