Question

If you draw a card 5 times (with replacement), what's the probability that you'll get exactly 1 ace?

Answer 1

with replacement means that all of your events are identically distributed. for instance, you need p(tffff) or p(ftfff) or ... 5 times. t means true for ace, f means false. Thus, you will need 5P(get an ace)=5 x 4/ 52 (or whatever the standard amount of cards is.

Answer 2

Jelly bean game =) 4 aces 52 cards, each time you take a jelly bean the percentage over the next goes higher. so for example 4/52 for the first card, and 4 out of 51 for the second. not sure if this is helpful or not =)

Answer 3

I would approach it this way:
To get exactly 1 ace, means that on any given draw, you get an ace, but then you can NOT get an ace on the other draws [and.. with replacement means that the card you draw goes back in the pile, so the deck you draw from on each turn is always the same (same # of cards)]

Furthermore, it doesn't specify on which turn you can get an ace, so it could happen on the 1st draw, OR the 2nd draw, OR the 3rd, OR the 4th, OR the 5.

So,
\(P(\text{draw exactly }1\text{ ace})=P(\text{ace 1st draw OR ace 2nd draw OR ... OR ace 5th draw)}\\
=P(\text{ace 1st draw})+\cdots +P(\text{ace 5th draw)}\) since the events are mutually exclusive.

Then, \[P(\text{ace 1st draw})=\frac{4}{52}\times \left(\frac{48}{52}\right)^4\], since the 1st draw must be an ace (with probability 4/52) and the other 4 draws cannot be an ace (with probability 48/52)

The same logic applies to P(ace 2nd draw) [1st draw will be a non-ace, 2nd draw is an ace, and the other 3 draws are non-ace] ... same logic until P(ace 5th draw).
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Another way to view this is using the binomial distribution. If we let \(X\) be the number of ace cards that show up, then the probability of this success (of an ace showing up) is 4/52 and there are 5 trials (i.e.5 draws)
\[ P(X=1)={5\choose 1}\left(\frac{4}{52} \right)^1 \left(1-\frac{4}{52}\right)^{5-1}\]