Two cards are drawn from an deck of 52 cards, without replacement. How do you find the probability that exactly one card is a spade?

Nov 22, 2016

The reduced fraction is $\frac{13}{34}$.

Explanation:

Let ${S}_{n}$ be the event that card $n$ is a spade. Then $\neg {S}_{n}$ is the event that card $n$ is not a spade.

$\text{Pr(exactly 1 spade)}$
$= \text{Pr"(S_1)*"Pr"(notS_2|S_1)+"Pr"(notS_1)*"Pr} \left({S}_{2} | \neg {S}_{1}\right)$

$= \frac{13}{52} \cdot \frac{39}{51} + \frac{39}{52} \cdot \frac{13}{51}$
$= 2 \cdot \frac{1}{4} \cdot \frac{39}{51}$
$= \frac{39}{102} = \frac{13}{34}$

Alternatively,

$\text{Pr(exactly 1 spade)}$
$= 1 - \left[\text{Pr(both are spades)"+"Pr(neither are spades)}\right]$
$= 1 - \left[\left(\frac{13}{52} \cdot \frac{12}{51}\right) + \left(\frac{39}{52} \cdot \frac{38}{51}\right)\right]$
$= 1 - \left[\frac{1}{4} \cdot \frac{12}{51} + \frac{3}{4} \cdot \frac{38}{51}\right]$
$= 1 - \left[\frac{12 + 114}{204}\right]$
$= 1 - \frac{126}{204}$
$= \frac{78}{204} = \frac{13}{34}$

We could also look at it as

$\left(\left(\text{ways to draw 1 spade")*("ways to draw 1 non-spade"))/(("ways to draw any 2 cards}\right)\right)$

$= \left({\text{_13"C"_1*""_39"C"_1)/(""_52"C}}_{2}\right)$

=((13!)/(12!1!)*(39!)/(38!1!))/((52!)/(50!2!))

$= \frac{13 \cdot 39}{\left(52 \cdot 51\right) / 2}$

$= \left({\cancel{2}}_{1} \cdot {\cancel{13}}^{1} \cdot {\text{^13cancel(39))/(cancel(52)_2^(cancel(4))*}}^{17} \cancel{51}\right)$

$= \frac{13}{34}$

This last way is probably my favourite. It works for any group of items (like cards) that have subgroups (like suits), as long as the numbers left of the C's on top $\left(13 + 39\right)$ add to the number left of the C on bottom $\left(52\right)$, and same for the numbers right of the C's $\left(1 + 1 = 2\right)$.

Bonus example:

What is the probability of randomly picking 3 boys and 2 girls for a committee, out of a classroom with 15 boys and 14 girls?

Answer: $\left({\text{_15"C"_3*""_14"C"_2)/(""_29"C}}_{5}\right)$