# Thread: hsc 2016 probability question

1. ## hsc 2016 probability question

Could someone explain this? I had a look at the answers and don't entirely understand it.. especially part 2.

from 11 f)
A darts player calculates that when she aims for the bullseye the probability of 3/5 her hitting the bullseye is with each throw.
(i) Find the probability that she hits the bullseye with exactly one of her first three throws.

(ii) Find the probability that she hits the bullseye with at least two of her first six throws.

2. ## Re: hsc 2016 probability question

i) This means that she hits a bullseye with one and misses it with her other two. There are three ways this can happen (she can hit it with either her first, second or third shot) and the probability of each is $3/5 \cdot (2/5)^2$.

ii) Method 2:

Hitting at least two bullseyes is the complementary event of hitting zero or one bullseye. Either she hits 0 or 1 bullseyes, or she hits 2-6 bullseyes.

There is only ONE way she can hit no bullseyes, and the probability of it is $(2/5)^6$.

There are $^6C_1 = 6$ ways she can hit one bullseye (you're choosing exactly ONE shot out of six to be the one that hits the bullseye) and the probability of each is $(3/5)(2/5)^5$.

Add these together and subtract from one to get the answer.

(The other solution in Method 1 is the long way - adding up the probabilities of hitting two bullseyes, three bullseyes, four bullseyes ... )

3. ## Re: hsc 2016 probability question

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