Probability of 4 successes of 7 tries @ 80% success rate?

This uses the Binomial distribution:

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

n : number of trials

k : number of successes

p : probability of success

q : probability of failure (1 - q)

In your case, you have:

n = 7 tries

k = 4 successes

p = 0.80

q = 1 - 0.80 = 0.20

And C(n,k) is the "n choose k" formula:

C(n,k) = n! / ((n-k)! k!)

We could plug in the numbers and come out with the answer, but let me help you see why the formula makes sense.

First imagine someone asked you to roll a die and they wanted to know the probability of getting the first 4 rolls as successes (0.8^4) and the last 3 rolls as failures (0.2^3).

The probability of that is:

0.8^4 * 0.2^3

But, that's only one way you could have exactly 4 successes. We asked for all the successes to happen on the first 4 rolls, but there are actually "7 choose 4" ways to pick the order of successes and failures and that would still be 4 successes and 3 failures. So we need to multiply the result above by C(7,4)

C(7,4) = 7! (4! 3!)

= 7 x 6 x 5 x 4 / (4 x 3 x 2 x 1)

= 7 x 6 x 5 / (3 x 2 x 1)

= 7 x 6 x 5 / 6

= 7 x 5

= 35 ways

The final equation:

P(X = 4) = C(7,4) * 0.8^4 * 0.2^3

= 35 * 0.8^4 * 0.2^3

= 1792 / 15265

≈ 0.114688

Answer:

About 11.47%

Update:

If you want the probability of *at least* 4 successes, then you can figure out P(X=4), P(X=5), P(X=6), P(X=7) and add them up. The result comes out to 0.1147 + 0.0852 or about 96.7% I'm not sure which calculator you have and what functions specifically it might have for doing this. WolframAlpha can give you the answer as shown below. But it is always better to know how to get it yourself.

No, 4/7 = 57%

Exactly 4 or at least 4?