Probability?

Question

Select one card randomly from a standard deck of 52 cards.
Let R be the event that the card is Red.
Let A be the event that the card is an ace.

Find:

(a) P(A and R)

(b) P(A/R)

(C) P(A or R)

Please explain how you got the answer. Thank You

Answer 1

OK.

There are 52 cards.

a) there are 2 red aces in the deck.
Hence P(A and R) = 2/52 = 1/26

b) Does P(A/R) mean the probability that if the drawn card is red, that it also is an ace?
Well there are 26 red cards in the deck. 2 of them are aces. So
P(A/R) = 2/26 = 1/13

c) P(A or R)
here we simply sum the probabilities P(A) and P(R) but have to subtract the possibilities we counted twice: i e cards that are aces AND red were counted twice. So

P(A or R) = P(A) + P(R) - P(A and R)

P(A) = 4/52 (4 aces in a deck)
P(R) = 1/2 (half of the deck is red)

P(A or R) = P(A) + P(R) - P(A and R) = 4/52 + 1/2 - 1/26
P(A or R) = (2 + 13 - 1)/26
P(A or R) = 14/26
P(A or R) = 7/13

Hope this helps.

Answer 2

a. p= 1/26. There are two red aces.
b. p=1/13. If a card is red, there is a 1 in 13 chance it is an ace.
c. p= 7/13. There are 4 aces (of which 2 are red) and 24 red cards for the other ranks. Thus you have 28 of 52 chances.

Answer 3

(a) there are two cards that are Red and Ace, therefore 2/52 or 1/16.

(b) I dont understand this. Do you mean, A, not Red? In that case it would also be 1/26 as there are two cards that are Aces and not Red.

(c) There are 26 Reds, plus two black Aces makes 28 cards that are Red or an Ace - therefore 28/52 or 7/13

Answer 4

a) There are 2 red aces in a deck, so the probability that you draw either one is:
P(A and R) = 2/52 = 1/26

b) Half of the deck are red, which is 26. So the probability that you draw a red ace when already know that it's a red card is:
P(A/R) = 2/26 = 1/13

c) P(A or R) is the probability that the card you draw is either an ace or a red one. So it is P(A), probability that it's an ace (including 2 red aces) plus P(R), probability that it's a red card (also including 2 red aces). Notice we count those 2 red aces twice, so we must subtract 1 time P(A)P(R), probability of drawing a red ace card:
P(A) + P(R) - P(A)P(R) = 4/52 + 26/52 - (1/13)(26/52) = 7/13

Answer 5

a. 2/52 (There are two red aces in the deck, and 52 cards altogether.)

b. I don't understand the question....

c. 28 (There are 26 red cards in the deck and 2 black aces in the deck, so there are 28 chances that the card you draw will be either an A or a red card.)

Answer 6

Hi, There are formulas you can see:
P(A and R)=P(A)+P(R)-P(AorR) so that
16 +13 -29 =0
16 is because you have 16 ace cards in all, 13 you have as many red and 29 is it together and this event mutually exclusive and equal to 0
P(A/R)=16/13 because R is you given, meaning that it 13
The last one you have to know from the first one

Answer 7

You need to know:

There are 4 of each type of card (i.e. ace, king queen, ten, nine, etc.) and there are 26 black cards and 26 red cards

a.) P(4 and 26)

b.) P(4/26)

c.) P(4 or 26)

Answer 8

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