Question

If two cards are drawn without replacement from a standard 52 deck. Find the probability that we draw at least one black card

Answer 1

Would the answer then just be \[\frac{ 26 }{ 52 }\]

Answer 2

Best and easiest way is to get the probability of "no black cards" and then subtract that from 1.

Answer 3

So, P("no black cards") = first one red x second one red.

Answer 4

BTW, 26/52 is not the answer.

Answer 5

Start by finding out what the P("first card red") is.

Answer 6

P(first card red) =1-\[\frac{ 26 }{ 52 }\]* \[\frac{ 26 }{ 51 }\]

Answer 7

P("first card red") = 26/52. Now, there are only 51 cards left of which 25 are red, so the P("first AND second card are red") = (26/52) (25/51). The P of your question is 1 - (26/52) (25/51).

Answer 8

That will cover the situation of: first card black, second red, also first card red, second black, also both black. So, instead of figuring all these three out, you just have to figure "both red" and subtract from one.

Answer 9

so \[\frac{ 1 }{ 2 } *\frac{ 25 }{ 51 } = \frac{ 25 }{ 102 } = 0.25 Therefore, 1-0.25 = 0.75\]

Answer 10

That's some pretty heavy-duty rounding, once you get to 0.25. I'd just leave the answer as 1-(25/105) which is closer to 0.755 Depends on how much liberty you were given for the rounding.

Answer 11

Wait where did you get the 105?

Answer 12

Sorry, typo. Meant to say 1-(25/102).

Answer 13

oh ok!

Answer 14

It's the "without replacement" that gives this problem its edge and funny second denominator.

Answer 15

ok this may sound silly but how would I write 1-\[\frac{ 25 }{ 102 }\]in simplified fraction form

Answer 16

np 1- (25/102) = (102 - 25)/102 = 77/102

Answer 17

Thank you so much!!