Question

A card is chosen at random from a pack of cards and replaced. This experiment is carried out 520 times. State the expected number of times on which the card is either an ace or a club or both.

Answer 1

1) P(a club) = 13/52
E( a club) = np = 520*13/52 = 130
2) P(an ace) = 4/52
E(an ace) = 520*4/52 = 40
3) P(a picture card) = 12/52
E(a picture card) = 520*12/52 = 120
4) P(either an ace or a club or both) = 4/52 + 13/52 - 1/52 + 1/52 = 17/52
E(either an ace or a club or both) = 520*17/52 = 170
5) P(neither an ace nor a club) = 1 - 16/52 = 36/52
E(neither an ace nor a club) = 520*36/52 = 360

Answer 2

The answer is 160... not 170... Not sure how to arrive at that answer though.

Answer 3

But, 360 is correct

Answer 4

probability of ace,club or both is 16/52

for 520 draws number times = 16/52 * v 520 = 160

Answer 5

How did you arrive at that answer?

Answer 6

well there are 13 clubs and 3 aces (excluding 3 clubs) total 16
on one draw therefore probabily of drawing 1 of these is 16/52

times 520 = 160

Answer 7

P(club) = 13/52
P(ace) = 4/52
P(club and ace) = 1/52
P(club or ace) = P(club)+P(ace)-P(club and ace) = 16/52

Answer 8

I thought there were 4 aces?

Answer 9

yes bu one is a club - so you dont count that twice

Answer 10

just remember the rule P(A or B) =P(A)+P(B)-P(A and B)

Answer 11

yep - trust us order!!

Answer 12

I am. And, it's right what you're writing... it's just you first said there are 3 aces... and I didn't see the exlusion part. Sorry about that. I understand it now :)

Answer 13

thats ok