# Thread: Binomial Distribution

1. ## Binomial Distribution

Alex has 5 shots at a dart board. He knows the probability of getting a bullseye is 0.6.

a) Find the probability that

i) No bullseye is hit. I get $\displaystyle P(X=0) = ^4C_0\times (0.6)^0(0.4)^5$

ii) At least one bulleyes is hit. I get $\displaystyle P(X\geq 1) = 1-^4C_0\times (0.6)^0(0.4)^5$

Here's where the confusion sets in.

b) Find the number of throws required such that the probability of at least one bullseye being thrown is at least 0.7

I know I have to make something >0.7.

2. Originally Posted by Bushy
Alex has 5 shots at a dart board. He knows the probability of getting a bullseye is 0.6.

a) Find the probability that

i) No bullseye is hit. I get $\displaystyle P(X=0) = ^4C_0\times (0.6)^0(0.4)^5$

ii) At least one bulleyes is hit. I get $\displaystyle P(X\geq 1) = 1-^4C_0\times (0.6)^0(0.4)^5$

Here's where the confusion sets in.

b) Find the number of throws required such that the probability of at least one bullseye being thrown is at least 0.7

I know I have to make something >0.7.
Just rework part ii for n throws instead of 5, then find the least n such that the probability is at least 0.7.

You should have $\displaystyle ^5C_0$ in your formulas instead of $\displaystyle ^4C_0$, but it won't make any difference in the final result.

You are implicitly assuming that the chances of hitting on different throws are independent. It would be good to state that assumption explicitly.