vcaa 2008 conditional probability

[captain_kirk]

Sharelle is the goal shooter for her netball team. During her matches, she has many attempts at scoring a
goal.
Assume that each attempt at scoring a goal is independent of any other attempt. In the long term, her scoring
rate has been shown to be 80% (that is, 8 out of 10 attempts to score a goal are successful).

What is the probability, correct to four decimal places, that her first 8 attempts at scoring a goal in a
match are successful?
= 0.1678

What is the probability, correct to four decimal places, that exactly 6 of her first 8 attempts at scoring
a goal in a match are successful?
= 0.2936

What is the probability, correct to three decimal places, that her fi rst 4 attempts at scoring a goal are
successful, given that exactly 6 of her fi rst 8 attempts at scoring a goal in a match are successful?



how do you work out the Pr(her fi rst 4 attempts at scoring a goal are successful|exactly 6 of her fi rst 8 attempts at scoring a goal in a match are successful)
the vcaa an itute notes don't denote any applicable formula so i'm still unsure as to how to determine Pr(A intersection B)
any help will be greatly appreciated

[skorm123]

I'm going to give this a try not sure if it is right but the trick is to find the intersection. In this case the top of the 'given formula' is Pr ( first she gets 4 out of 8 attempts) is the intersection of
 Pr(first 4 out of 8 attempts) intersect Pr(first 6 out 8 attempts). Therefore this can be substituted in for 0.046/0.29 =0.16

[illuminati]

nah its just 0.1678/0.2936

[captain_kirk]

the working is (0.8^4 x binompdf(4, 0.8, 2))/0.2936 = 0.214

i don't know how they determine the top of the fraction

[illuminati]

oh nvm haha
the top fraction is the intersection of getting 4 of your first ones and then 6 altogether
So the 4 first shots is 0.8^4
2 of the last 4 shots is 4C2 x 0.8^2 x 0.2^2
so like, 0.8^4 x 4C2 x0.8^2 x 0.2^2 is your fraction