Question

Phillip flips an unfair coin eight times. The coin is twice as likely to come up heads as tails. How many times as likely is Phillip to get exactly three heads than exactly two heads?

Answer 1

step one
if the coin is twice as likely to come up heads, what is the probability it comes up heads?

Answer 2

2/3 to come up with heads.

Answer 3

ok good
now we can use binomial probablility
\[P(x=k)=\binom{n}{k}p^k(1-p)^{n-k}\] with
\[n=8, k=2, p=\frac{2}{3}, 1-p=\frac{1}{3}\]

Answer 4

if this is greek i can explain it more slowly but it may look familiar

Answer 5

I'm still following

Answer 6

ok so lets compute the probability you get exactly 3 heads
it is
\[P(x=3)=\binom{8}{3}(\frac{2}{3})^3(\frac{1}{3})^5\]

Answer 7

a little calculator help gives me
\[\frac{448}{6561}\]
http://www.wolframalpha.com/input/?i=%288+choose+3%29%282%2F3%29^3%281%2F3%29^5

Answer 8

oh no that is not the question

Answer 9

the question is
"How many times as likely is Phillip to get exactly three heads than exactly two heads?"

Answer 10

so we still have to compute
\[P(x=2)\]

Answer 11

let me know when you get that one

Answer 12

\[\frac{ 112 }{ 6561 }\]

Answer 13

then divide the first one by the second one to get your answer

Answer 14

4

Answer 15

pretty clear that the answer is 4,
and it would probably have been easier if we didn’t compute anything, just wrote the two formulas down and cancelled our brains out

Answer 16

but that is by hindsight and i have no desire to write it down and see
you might though

Answer 17

Something like this? http://gyazo.com/529796e0aeadc0ec583e4fd7fbaaaa90

Answer 18

\[\frac{\dbinom{8}{3}\left(\frac{2}{3}\right)^3\left(\frac{1}{3}\right)^5}{\dbinom{8}{2}\left(\frac{2}{3}\right)^2\left(\frac{1}{3}\right)^6}\]

Answer 19

look at all the common factors!

Answer 20

\[\frac{\binom{8}{3}}{\binom{8}{2}}=2\]

Answer 21

and also \[\frac{\frac{2}{3}}{\frac{1}{3}}=2\]

Answer 22

impress your teacher if you have to hand it in
do it the smart way