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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.

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    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. – 2017-01-02
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    Add it to the question please. – 2017-01-02
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    BTW, the answer is $\frac{\binom{4}{1}\cdot\binom{40-4}{2-1}}{\binom{40}{2}}$, can you see why? – 2017-01-02
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    I remember to multiply but i never used 36, the 40-4 i mistakenly always used 4/ 40 then later 1/10 – 2017-01-02

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Hint:

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

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    Why is that "$2!$"? – 2017-01-02
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    where does $2!$ come from?? – 2017-01-02
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    It doesn't, according to the title (and according to common sense). – 2017-01-02
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    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... – 2017-01-02
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    I don't agree with the above comments. Whenever a particular order for a hypergeometric distribution isn't specified, order **is** important; using combinations automatically takes care of this, while for direct multiplication of probabilities, a multiplier (here, $2!$) will be needed. – 2017-01-02
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    @trueblueanil: The title doesn't say "first ball is an ace". It says "1 ace and 1 not ace". So I think that's pretty much outright "order is NOT important". – 2017-01-02
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    @trueblueanil I also agree with you. – 2017-01-02
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    In either case, considering the order when counting the number of desired cases, but not when counting the total number of cases (i.e., multiplying by $2!$ the numerator but not the denominator), is wrong, whether the order is important or not important. But I see that you've fixed that, so no point arguing about it any further... – 2017-01-02
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    @barakmanos: I prefer to use the term "order **matters**" for clarity. Since in the question it hasn't been specified that the first card is an ace, order **does** matter, but whether you need to multiply by $2!$ or not depends upon the way you compute the result, as pointed out in my answer. – 2017-01-02
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    @trueblueanil I tend to consider order by default except if it is said "without order" explicitly (+1). – 2017-01-02
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    @msm: I am correcting the downvote (given by someone else earlier) because the answer as it stands is correct. – 2017-01-02
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    @trueblueanil: I've been thinking about it. In general could say that the order always matter. It's just that sometimes, all orders are equally matter (which is equivalent to what we often refer to as "the order does not matter"). – 2017-01-02
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    @barakmanos: By now, I'm also confused as to what "order is important/matters" means ! May be, as I wrote in my answer, it is better to write "when a specific order isn't specified, all orders have to be considered." – 2017-01-02
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    @trueblueanil: As I said, I believe that the terms `order is not important` and `all orders are equally important` are **equivalent**. So you could either solve a given question using combinations (for the first case) or permutations (for the second case). BTW, it might be worth posting here a question on whether or not these terms are indeed equivalent. – 2017-01-02
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    @barakmanos: I have posted it at http://math.stackexchange.com/questions/2080620/semantic-confusion-over-the-meaning-of-order-matters-is-important – 2017-01-02
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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 $\dfrac{\binom41\binom{36}1}{\binom{40}2}$

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    Can you please give more details about your answer? – 2018-03-19
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    @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 $\frac4{40}\times \frac{36}{39}$ – 2018-03-24
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    I understand this part, I don't understant why is it multiplied by 2! – 2018-03-24
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    ok, it is derived by solving using combinations – 2018-03-24
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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 = $\frac{4}{40} \cdot \frac{36}{39} + \frac{36}{40} \cdot \frac{4}{39}$

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    It seems that you also agree that ordering matters +1 – 2017-01-02
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    @msm Actually, both cases being considered to be the same event, means that "order does *not* matter". – 2017-01-02
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    I agree @GrahamKemp. In this particular case, please notice the terms "first" and "second" in the post which imply order. – 2017-01-02
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    Yes I agree with @msm. – 2017-01-02
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    http://www.math.uiuc.edu/~wgreen4/Math124S07/PermuCombinations.pdf – 2017-01-02