$40$ 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. – dbz.blue Jan 2 '17 at 8:50
• Add it to the question please. – barak manos Jan 2 '17 at 8:51
• BTW, the answer is $\frac{\binom{4}{1}\cdot\binom{40-4}{2-1}}{\binom{40}{2}}$, can you see why? – barak manos Jan 2 '17 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 – dbz.blue Jan 2 '17 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 $\dfrac{\binom41\binom{36}1}{\binom{40}2}$

• Can you please give more details about your answer? – Andrei Mar 19 '18 at 17:19
• @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 $\frac4{40}\times \frac{36}{39}$ – true blue anil Mar 24 '18 at 13:56
• I understand this part, I don't understant why is it multiplied by 2! – Andrei Mar 24 '18 at 13:57
• ok, it is derived by solving using combinations – Andrei Mar 24 '18 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 Jan 2 '17 at 9:56
• @msm Actually, both cases being considered to be the same event, means that "order does not matter". – Graham Kemp Jan 2 '17 at 10:22
• I agree @GrahamKemp. In this particular case, please notice the terms "first" and "second" in the post which imply order. – msm Jan 2 '17 at 10:30
• Yes I agree with @msm. – Kanwaljit Singh Jan 2 '17 at 10:36
• math.uiuc.edu/~wgreen4/Math124S07/PermuCombinations.pdf – true blue anil Jan 2 '17 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 "$2!$"? – barak manos Jan 2 '17 at 8:54
• where does $2!$ come from?? – user394255 Jan 2 '17 at 8:55
• It doesn't, according to the title (and according to common sense). – barak manos Jan 2 '17 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... – barak manos Jan 2 '17 at 8:58
• @barakmanos: I have posted it at math.stackexchange.com/questions/2080620/… – true blue anil Jan 2 '17 at 15:41