# Data Management: Binomial Distributions?

Find the probability that in 10 tosses of a fair coin there will be at least 2 years. Please show how you got your answer.

Sorry, I meant heads, no years. Stupid autocorrect :/

*not

### 2 Answers

- PuzzlingLv 71 year agoFavorite Answer
I'll assume you meant at least 2 *heads*.

The binomial probability formula is:

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

n : number of trials (10)

k : number of successes

p : probability of success (1/2)

q : probability of failure (1/2)

You want the probability that the number of heads is 2 or more. So you could figure out:

P(X ≥ 2) = P(X = 2) + P(X = 3) + P(X = 4) + ... + P(X = 10)

But that's a lot of computation. So instead, let's figure the *opposite* probability and later we can subtract from 1.

In other words, start with the probability that you get 0 heads or 1 head. That way you only have to do 2 calculations instead of 9.

P(X = 0) = C(10,0) * ½^0 * ½^10

= 1 * ½^10

= 1/1024

P(X = 1) = C(10,1) * ½^1 * ½^9

= 10 * ½^10

= 10/1024

Adding those we get:

P(X < 2) = P(X = 0) + P(X = 1)

= 1/1024 + 10/1024

= 11/1024

But you want the opposite case, so subtract from 1:

P(X ≥ 2) = 1 - P(X < 2)

= 1 - 11/1024

= 1024/1024 - 11/1024

= 1013/1024

Answer:

1013/1024 ≈ 98.926%

P.S. And if you actually meant at least two *tails*, the answer is still the same by symmetry (both have the same probability).

My first thought was a probability of zero since I foresee no circumstances in which it would take 2 or more years to toss a coin 10 times. But your answer is probably closer to what was intended.