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**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.