probability of drawing cards

[Clifford]

I don't think I am on the right track with these probability questions. Could somebody give me a hand?

Two cards are drawn from a given a standard deck of 52 cards, find the probabiliy of each event:
a) both cards are diamonds
b) both cards are red
c) both cards at 7's
d) both cards are red 7's
e) both are 7's or both are red

a) there are 13 diamonds, 1/4 to draw the first diamond and 12/51 to draw the 2nd.
1/4 * 12/51 = 1/17

b) 1/2 * 25/51 = 25/102

c) 1/13 * 3/51 = 1/221

d) 1/26 * 1 / 51 = 1/1326

e) 25/102 * 1/221 = 331/1326

 

[pka]

Here are two answers. You should use them to work out the others.
\(\displaystyle \L \begin{array}{l}
b)\quad \frac{{\left( {\begin{array}{c}
{26} \\
2 \\
\end{array}} \right)}}{{\left( {\begin{array}{c}
{52} \\
2 \\
\end{array}} \right)}} \\
d)\quad \frac{1}{{\left( {\begin{array}{c}
{52} \\
2 \\
\end{array}} \right)}} \\
\end{array}
.\)

 

[Clifford]

does this look better?

a) C(13,2) / C(52,2)

b) C(26,2) / C(52,2)

c) C(4,2) / C(52,2)

d) 1 / C(52,2)

e) C(26,2) / C(52,2) * C(4,2) / C(52,2)

 

[pka]

Much better!
However, for the last one did you forget inclusion/exclusion?
\(\displaystyle \L \left| {S \cup R} \right| = \left| S \right| + \left| R \right| - \left| {S \cap R} \right|\).

 

[soroban]

Hello, Clifford!

There was nothing wrong with the way you wrote it.
Your reasoning and your work are correct . . . except the last one.

Two cards are drawn from a given a standard deck of 52 cards.
Find the probabiliy of each event:

a) both cards are diamonds

There are 13 diamonds.
13/52 to draw the first diamond and 12/51 to draw the 2nd.
. . \(\displaystyle P(\text{two diamonds})\;=\,\frac{13}{52}\,\cdot\,\frac{12}{51}\;=\;\frac{1}{17}\)


b) both cards are red
26/52 to draw the first red and 25/51 to draw the second.
. . \(\displaystyle P(\text{two Red})\;=\;\frac{26}{52}\,\cdot\,\frac{25}{51}\:=\:\frac{25}{102}\)


c) both cards at 7's
4/52 to draw the first 7 and 3/51 to draw the second.
. . \(\displaystyle P(\text{two 7's})\;=\;\frac{4}{52}\,\cdot\,\frac{3}{51}\;=\;\frac{1}{221}\)


d) both cards are red 7's
2/52 to draw the first red 7 and 1/51 to draw the second.
. . \(\displaystyle P(\text{two red 7's}) \;=\;\frac{2}{52}\,\cdot\,\frac{1}{51}\;=\;\frac{1}{1326}\)


e) both are 7's or both are red

\(\displaystyle P(\text{two 7's}\) or \(\displaystyle \text{two Reds})\;=\;P(\text{two 7's})\,+\,P(\text{two Reds)\,-\,P(\text{two red 7's})\)

. . . . . . . . . . . . . . . . . . \(\displaystyle =\;\L\frac{1}{221}\,+\,\frac{25}{102} \,-\,\frac{1}{1326} \;=\;\frac{55}{221}\)