[Math] $40$ cards, $4$ are aces. Probability of $1$ ace when drawing $2$ cards at ramdom

probability

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.

Best Answer

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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