Probability of drawing a card.?

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The probability of drawing a desired card on a single draw is:
4/5 = 80%

The probability of drawing a desired card every time on two draws is:
4/5 x 4/5 = 16/25 = 64%

The probability of drawing a desired card every time for *twelve* draws is:
4/5 x 4/5 x 4/5 x ... x 4/5
= (4/5)^12
≈ 6.9%

Now if you are talking about some of the trials, say 4 out of the 12 trials you have to get into combinations. First, let's figure the probability that the first 4 draws are the desired card:
4/5 x 4/5 x 4/5 x 4/5

Then let's make the other 8 draws not be the card:
1/5 x 1/5 x 1/15 x 1/5 x 1/5 x 1/5 x 1/5 x 1/5

So the total probability is:
(4/5)^4 x (1/5)^8

But this was only for one specific way of drawing 4 good cards and 8 bad cards. There are actually numerous ways to do that. There are "12 choose 4" ways to do that.

C(12,4) = 12! / (12-4)! 4!
= (12 x 11 x 10 x 9) / (4 x 3 x 2 x 1)
= 495 ways

So multiply the single probability by 495.
495 x (4/5)^4 x (1/5)^8
So the probability that exactly 4 of the trials would be good is:
≈ 0.052%

Here's the general formula:
p is the probability of the event happening on a single trial
q is the probability of it not happening

n is the number of trials
k is the exact number of trials where you'll have success.

P(n = k) = C(n,k) * p^k * q^(n-k)

For your case:
p = X/Y
q = 1 - X/Y
n = W
k = Z

If you are interested, read more about binomial probability....Show more