There is a question on section b (question 8) about megan throwing darts at a dartboard. I worked it out using the binomial distribution formula (ncr) and got 0.6484 or something. However in the question, they say 'you pay use (p+q)^3 as' bla blah, and wondering what relevelance that has to the question- it didnt help me! does that mean the answer I have is wrong? thanks!
Aw I was asking my Maths teacher for a copy of it since we did the blasted thing and I never got one
aww that sucks! lmao Well ive come to the conclusion that this exam wnats a different method (the cgp way). Its nice that they give you the expansion of (p+q)^3- although i cant for the life of me work it out (i understand (p+q)^2 though!) but nevermind. So if the formula is
p^3 + 3p^2q+3pq^2+q^3, do I just have to substitute p with 7/9, and q with 2/9? Or do I just change them to fractions or what? ahhh..
There is a question on section b (question 8) about megan throwing darts at a dartboard. I worked it out using the binomial distribution formula (ncr) and got 0.6484 or something. However in the question, they say 'you pay use (p+q)^3 as' bla blah, and wondering what relevelance that has to the question- it didnt help me! does that mean the answer I have is wrong? thanks!
EDIT: Scroll down to my 3rd post for more :P
Megan throws darts at a target. She can hit the target with 7 out of 9 throws. Megan throws three darts.
Assuming a binomial distribution, work out the probability that she hits the target exactly twice. You may use (p+q)^3, which is equal to: p^3 +3p^2q+3pq^2+q^3.
Ok, firstly you need to know that p=success outcomes (7/9)and q=unsuccessful outcomes (2/9)
So, we are looking for the part of the formula where P is squared and that part can only be 3p^2q Therefore: substituting the values we have; 3*7/9^2 *2/9 This then is equal to 98/243, which in decimal form is 0.403