Another probability Question.?

There are 13 cards in each suit

There are 4 aces in the deck

So it would seem that there might be 17 cards to count, except one of those is the ace of diamonds which would have been counted twice. Subtract 1 for that.

In total there 16 cards that are diamonds or aces out of 52.

P(diamond or ace) = 16/52

That can be reduced by dividing top and bottom by 4:

Answer:

4/13

P(getting a diamond) = P(A) = 13/52

P(getting an ace) = P(B) = 4/52

P(A and B) = 1/52

According to the Addition Theorem of Probability

P(getting a diamond or an ace) = P(A or B)

= P(A) + P(B) - P(A and B) --------> [because the events A and B are NOT mutually exclusive]

= 13/52 + 4/52 - 1/52

= 16/52 or 4/13

What is this? F*cking kindergarden math class? Stop jerking off and actually study for once. See this is what is wrong with this country, besides the jews, is lazy pieces of **** like you. Why don't you pay attention in school rather than sticking your d*ck in a crusty tube sock and wanking off to trannys or whatever f*cked up **** you're probably into, you f*cking piece of sh*t. I hope you die.