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Zielflug [23.3K]
3 years ago
11

You have a standard deck of cards. The deck has 52 total cards and contains 4 suits: hearts, clubs, diamonds, and

Mathematics
2 answers:
umka21 [38]3 years ago
7 0

Answer:

P(A)=0.25

P(B)=0.077

P(AB)=0.192

P(BA)=0.083

Check below for the full numbers, I rounded it for you.

Step-by-step explanation:

What is P(A), the probability that the card is a club?

So whenever you get a deck of cards you'll have 52 which have 4 suits, think of it like UNO, and club is one of the suit's name! This mean all this question is asking you is the probability of getting a club out of the 52 cards which is

52/4=13

So we know that there is 13 club cards.

Meaning the probability is a 25% since you can only get it 1/4 of the time.

P(A)=0.25

What is P(B), the probability that the card is a 5?

Same as what I stated earlier but this time you only have 4 cards that can be a "5" out of the 52 cards.

4/52=0.076923

P(B)=0.077

What is P(A and B), the probability that the card is a club and a 5?

Keep a eye on the working "and" this means it's referring to the possibility of getting a club and a 5 and since this can only happen once the chances are extremely slim.

1/52=0.1923076

P(AB)=0.192

What is P(BA), the conditional probability that the card is a 5 given that it is a club?

This assumes that you're only drawing from a deck of ONLY clubs so the chances of getting a 5 in that stack.

1/12=0.08333 (3 keeps going)

P(BA)=0.083

(This one might be wrong since I haven't done conditional probability in a long time but I feel somewhat confident, I do feel confident about the rest so no worries.)

Is P(BA) = P(B)? Are the events A and B independent?

No, P(BA) = P(B) is not true since A and B do influence one another since what if I get a 5 without replacement this results in change of chances of A.

Amiraneli [1.4K]3 years ago
7 0

Answer:

1.1/4

2.1/13

3.1/52

4.1/13

5. Yes p(B/A)=P(B) and Yes, events A and B are independent events.

Step-by-step explanation:

khan academy walk through

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