# $40$ cards, $4$ are aces. Probability of $1$ ace when drawing $2$ cards at ramdom?

In a deck of $40$ cards there are $4$ aces. What is the probability that when drawing two cards only one ace is drawn.

What I've come up with: $40$ cards $4/40$ are aces. Also chance first card is an ace (or $1/10$ simplified) If the first card is an ace the probability that the second card is an ace is: $3/39$ and if it isnt is $36/39$

I dont know or have and equation just numbers If a card is drawn and it ISNT an ace the chance the next card is an ace is $4/39$ and $35/39$ chance it isnt an ace.

• Ive got the fractions 4/40 and if the first card is an ace the the chance the second card isnt an ace is 36/39. Commented Jan 2, 2017 at 8:50
• Add it to the question please. Commented Jan 2, 2017 at 8:51
• BTW, the answer is $\frac{\binom{4}{1}\cdot\binom{40-4}{2-1}}{\binom{40}{2}}$, can you see why? Commented Jan 2, 2017 at 8:52
• I remember to multiply but i never used 36, the 40-4 i mistakenly always used 4/ 40 then later 1/10 Commented Jan 2, 2017 at 9:03

Since students are frequently confused about a "multiplier" when drawing w/o replacement, please note a few points

• when a specific order isn't specified, all orders have to be considered.

• if solving multiplying probabilities, you must therefore use a multiplier, viz. $\frac4{40}\cdot\frac{36}{39}\times 2!$

• if solving using combinations, all orders automatically get considered, thus $\dfrac{\binom41\binom{36}1}{\binom{40}2}$

• @Andrei: Look at another version to understand. If the question specifically mentioned that the first is an ace and the second isn't, the answer would simply be $\frac4{40}\times \frac{36}{39}$ Commented Mar 24, 2018 at 13:56
• I understand this part, I don't understant why is it multiplied by 2! Commented Mar 24, 2018 at 13:57
• ok, it is derived by solving using combinations Commented Mar 24, 2018 at 14:00

We have two cases.

First card is an ace and second other.

$\frac{4}{40} \cdot \frac{36}{39}$

First card is other and second is an ace.

$\frac{36}{40} \cdot \frac{4}{39}$

Total = $\frac{4}{40} \cdot \frac{36}{39} + \frac{36}{40} \cdot \frac{4}{39}$

• It seems that you also agree that ordering matters +1
– msm
Commented Jan 2, 2017 at 9:56
• @msm Actually, both cases being considered to be the same event, means that "order does not matter". Commented Jan 2, 2017 at 10:22
• I agree @GrahamKemp. In this particular case, please notice the terms "first" and "second" in the post which imply order.
– msm
Commented Jan 2, 2017 at 10:30
• Yes I agree with @msm. Commented Jan 2, 2017 at 10:36
• math.uiuc.edu/~wgreen4/Math124S07/PermuCombinations.pdf Commented Jan 2, 2017 at 11:42

Hint:

You can choose from $\binom{4}{1}\binom{36}{1}2!$ combinations where there are totally $\binom{40}{1}\binom{39}{1}$ options.

• Why is that "$2!$"? Commented Jan 2, 2017 at 8:54
• where does $2!$ come from??
– user394255
Commented Jan 2, 2017 at 8:55
• It doesn't, according to the title (and according to common sense). Commented Jan 2, 2017 at 8:56
• And even if we wanted to take the order into account, we would nevertheless need to permute (rearrange) both the numerator and the denominator. In other words, we'd need to multiply both parts by $2!$, or... we could simply reduce that factor because it has no effect at the bottom line... Commented Jan 2, 2017 at 8:58
• @barakmanos: I have posted it at math.stackexchange.com/questions/2080620/… Commented Jan 2, 2017 at 15:41