TO ALL STATISTICS BRAINS!!...how do u solve this???

Question

My question deals with binomial random variables.

The problem says that i need to find the probability that AT LEAST 1 ..bla bla bla.
Given: p=0.6 , n=3

I do not know how to simplify the Pr(x≥1) to get an answer using the binomial table.

Please i reaallly need help for this....

So is a basic rule of thumb:

Pr(X≥x) = 1 – Pr(X ≤ x-1)

Cause if it is, then I'll be ok!!

Answer 1

You can do this two ways...

1. Find P(x=1) + P(x=2) + P(x=3).

2. An easier way: find P(x >= 1) by taking P(x = 0) and subtracting that from 1.

Either way, from the table, you should get the same thing.

edit: That rule of thumb will work for the binomial distribution because it is discrete. It would not for, say, continuous distributions. A better rule of thumb might be this...

P(X >= x) = 1 - P(X < x).

This will always work. Hopefully you understand the reason behind it; the probability of an event is equal to one minus the probability of the opposite of that event.

Answer 2

P(x > = 1) = 1 - P(x = 0) you get the value and just subtract it from 1

Answer 3

p(x ≥ 1), given p(x) = 0.6, and n = 3
let p(NOTx) = q(x) = 1-p(x) = 0.4
then the probability of unsatisfactory results becomes q^1, q^2, q^3, . . . .q^n, so the probability of at least one desired event is:
p(x ≥ 1) = 1 - q^n
p(x ≥ 1) = 1 - 0.4^3
p(x ≥ 1) = 1 - 0.064
p(x ≥ 1) = 0.936