The question is as follows:
A certain deck of cards contains 2 blue cards, 2 red cards, 2 yellow cards, and 2 green cards. If two cards are randomly drawn from the deck, what is the probability that they will both are not blue?

My Solution was:

The probability of drawing a blue card on the first draw is

= (number of cards that are blue) / (total number of cards)
= 2/8

The probability of not drawing a blue card on the second draw is
= (number of cards that are blue left) / (total number of cards)
= 1/7
Probability of both cards being blue:
= (2/8) * (1/7)
= 1/28
Therefore probability of both cards NOT being blue= 1-(1/28)=27/28

But apparently the correct answer is:
The probability of not drawing a blue card on the first draw is

= (number of cards that are not blue) / (total number of cards)
= 6/8

The probability of not drawing a blue card on the second draw is
= (number of cards that are not blue) / (total number of cards)
= 5/7

= (6/8) * (5/7)
= 30/56
= 15/28

Shouldn't the two answers match? What is wrong with my reasoning here? Please help. Thanks!

Source of the above problem: The Delta Course (Probability Practice Questions#3)

I see why you are confused; these problems can be tough.

Your reverse probability solution is flawed because it does not take into account the probability that ONE of the cards might be blue.

You were working under the assumption that the opposite of neither card being blue is both cards being blue.

However, the opposite of neither card being blue is either a.) both cards are blue, or b.) one card is blue.

Hope that helps.

oh wow! thanx for clarifying that.

looks like this issue has already been resolved. just one general comment:
in probability problems with different cases (like this one), you must be very careful to make exhaustive lists of all possibilities, making sure to organize the possibilities in a way that ensures you won't leave some out.

in this problem, there are 4 distinct possibilities:
both cards blue
first card blue, second card not blue
first card not blue, second card blue
neither card blue

your first solution doesn't work because it neglects the second and third of these possibilities. had you taken a few seconds to organize (at least mentally, but better yet on paper) the different cases, you wouldn't have made this mistake.

happy hunting!

so true...thank you for the advice.

two cards are drawn without replacement from a standard pack of 52 playing cards:
there is two event A and B
A be event that the second card drawn is A
B be the event that the first card drawn is an A
what is the probability of P(A) and P(B)

whoa. you need to re-word this question please, as it's totally unclear what you're trying to ask. two big problems:

(1) i don't know what you mean by the A's: it looks like one A is an event, but the second and third A's are cards. do those A's stand for "ace"? if so, then write "ace." if not, then explain what cards you're talking about.

(2) what is the probability of P(A) and P(B)
i don't know whether this means
- two different values, the probability of 'a' and the probability of 'b', or
- one value, the probability of both 'a' and 'b' occurring.

please clarify. thanks