Adding Probability

I'm really stuck on this problem:

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Two cards are drawn from a deck of cards. What is the probability of having drawn a black card or an ace?

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Heres what I have so far:

- Drawing a black card and drawing the 2nd card being any other card: (26/52) * (51/51)

- Drawing an ace and drawing the 2nd card being any other card: (4/52) * (51/51)

- If there's a black ace: (2/52)


Well that's all I can think of. Ya, I sound really pathetic and really not knowing what I'm really doing :? Sorry

Thanks

Hello, AirForceOne!

You're expected to be familiar with this formula:

. . \(\displaystyle P(A\,\vee\,B)\:=\(A)\,+\,P(B)\,-\,P(A\,\wedge\,B)\)

It says "The probability of A or B happening is:
. . the probability of A plus the probability of B
. . minus the prob. of both A and B

Two cards are drawn from a deck of cards.
What is the probability of having drawn a black card or an ace?

Click to expand...

\(\displaystyle P(\text{black}) \:=\:\frac{26}{52}\)

\(\displaystyle P(\text{Ace}) \:=\:\frac{4}{52}\)

\(\displaystyle P(\text{black }\wedge\text{ Ace}) \:=\:\frac{2}{52}\)

Therefore: \(\displaystyle \(\text{black }\vee\text{ Ace}) \;=\;\frac{26}{52}\,+\,\frac{4}{52}\,-\,\frac{2}{52}\;=\;\frac{28}{52} \;=\;\L\frac{7}{13}\)

Hey I got that answer in one try, but the book says they answer is 175/221...
No wonder why I've been trying this problem for hours...I'll take your word for it.

Thanks!

What Soroban gave you was if one card is drawn.

For 2 cards, multiply 7/13 by

\(\displaystyle \L\\\frac{C(26,2)C(4,2)}{C(52,2)}\)

Just one way to look at it.

Or \(\displaystyle \L\\\frac{7}{13}(\frac{26}{51}+\frac{49}{51})=\frac{175}{221}\)

Or another way: \(\displaystyle \L1 - \frac{{\left( {\begin{array}{c}
{24} \\
2 \\
\end{array}} \right)}}{{\left( {\begin{array}{c}
{52} \\
2 \\
\end{array}} \right)}} = \frac{{175}}{{221}}.\)

That is the opposite of no black cards and no aces.

ahh I see.

Thanks a ton guys!

EDIT: Hey pka that's an interesting method. What algebra syntax is the 24 and the 2 on the bottom and the 52 and the 2 on the bottom?

And can you please explain the method to me? Thanks!

AirForceOne said:

What algebra syntax is the 24 and the 2 on the bottom and the 52 and the 2 on the bottom?

Click to expand...

The notation \(\displaystyle \left( {\begin{array}{c}
N \\
k \\
\end{array}} \right) = \frac{{N!}}{{k!\left( {N - k} \right)!}}\) is standard for the binominal coefficient or choosing k elements from N-element set.

As for the 24; remove all black cards and all aces then there are 24 cards left. Choose 2 from 24 and we have the opposite of what we want: one black or an ace.

There are \(\displaystyle {52} \choose 2\) to choose two cards from the deck.

pka said:

AirForceOne said:

What algebra syntax is the 24 and the 2 on the bottom and the 52 and the 2 on the bottom?

Click to expand...

The notation \(\displaystyle \left( {\begin{array}{c}
N \\
k \\
\end{array}} \right) = \frac{{N!}}{{k!\left( {N - k} \right)!}}\) is standard for the binominal coefficient or choosing k elements from N-element set.

As for the 24; remove all black cards and all aces then there are 24 cards left. Choose 2 from 24 and we have the opposite of what we want: one black or an ace.

There are \(\displaystyle {52} \choose 2\) to choose two cards from the deck.

Click to expand...

ahhh I see...I got confused with the syntax because I didn't know it's the same thing as writing C(24,2) etc...

Wow this method is so much easier and makes more sense!

Thanks!