Question

A card is drawn from a standard deck of cards. Determine whether the events are mutually exclusive or inclusive. Then, find the probability.

P(ace or club)

a. inclusive, 4/13
b. mutually exclusive, 17/52
c. mutually exclusive, 4/13
d. inclusive, 17/52


Can you explain this to me step by step ?
I actually want to know how to do it.

Answer 1

post all of your homework questions in one thread so people can feast on them with response. i don't think you're even trying to solve any of them by yourself.

Answer 2

i am trying, the ones i'm posting up are the ones i don't understand at all.

Answer 3

you know the formula for probablilty (what you want) / (total outcomes) yes?

Answer 4

no i don't even know how to start the problem

Answer 5

ok

Answer 6

probablilty is calculating the fraction of something you want to happen. in your case you want to draw either a ace or a club from a deck of cards. so start simpler, do you know the probablity of pulling a 4 from a deck of cards?

Answer 7

so that's 4/52 which will be 1/13?

Answer 8

yes good... what you have is a mutually exclusive event because you can have BOTH an ace and a club at the same time... so there's a special formula for that

Answer 9

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

Answer 10

so we'll say that there are 52 cards in the deck and that will be our denominator, we'll focus on the numerators. How many aces are in a deck of cards?

Answer 11

4

Answer 12

ok so P(A) = 4/52... how many clubs are in a deck of cards?

Answer 13

1/13?

Answer 14

no, there's 13 clubs in a deck of cards so P(B) = 13/52

Answer 15

now here's the tricky part of the problem is that there is a CHANCE of getting both AN ACE and A CLUB at the same time. that's what makes it mutually exclusive so you have to subtract this chance away so you don't count it twice. so P(A) = 4/52 P(B) = 13/52 and P(A and B) = 1/52.

So what is P(A) + P(B) - P(A or B)?

Answer 16

oops should be what is P(A) + P(B) - P(A and B)

Answer 17

P(4/52) + P(13/52) - P(4/52 and 13/52)
P(17/52) - P(4/52 and 13/52)
so on the P(A and B) part do i add them together?

Answer 18

no P(A and B) is the probablity of getting BOTH an ace and a club. how many aces of clubs are in a deck of cards?

Answer 19

13/52

Answer 20

remind me not to play cards with you :) there's 13 clubs in a deck of cards, how many of those clubs are aces also?

Answer 21

there's 13 clubs but 4 of them are aces

Answer 22

no there is only 1 ace of clubs elsa

Answer 23

there are 4 aces TOTAL in the deck, but not all of the aces are clubs

Answer 24

the probablilty of drawing an ace and a club at the same time is 1/52

Answer 25

P(A) = probality of drawing an ace = 4/52
P(B) = probability of drawing a club = 13/52
P(A and B) = probability of drawing both at the same time = 1/13
P(A or B) = P(A) + P(B) - P(A or B) = (4 + 13 - 1) / 52

Answer 26

grr last line should read:
P(A or B) = P(A) + P(B) - P(A and B) = (4 + 13 - 1) / 52