bayes theorm and conditional probability

Question

There are two coins on a table. When both are flipped, one coin lands

on heads with probability 0.5 while the other lands on heads with

probability 0.6. A coin is randomly selected from the table and flipped.

(a) What is the probability it lands on heads?

(b) Given that it lands on tails, what is the conditional probability that

it was the fair coin (that is, the one equally likely to land heads or

tails)?

Answer 1

(1/2)(.5)+(1/2)(.6)=.25+.30=.55

or just average .5 and .6. Midpoint is .55

You could get this by reading off the 2nd row of Pascal's Triangle, 1 1. Each 1 is 1/2 the total. 1/2 times .5 plus 1/2 times .6 = .55

That's the P(H) for one coin flip.

IF you flipped both coins, the probability of at least one head would be greater. Go to the next row of Pascal's Triangle 1 2 1 1 is 1/4 of the total 2 is 1/2 the total 1+2+1=4

You get 1 way to get 2 heads, HH, 2 ways to get 1 head and one tail, HT, TH. and one way to get zero heads: TT. 3/4 of the time you will get at least one head. P(at least one H/2 flips)=.75 if both coins had P(H)=1/2=.5 but one coin has P(H)=.6, so we'll get a P(at least one H/given 2 flips)>.75 but less than 1.00

Take the 3rd row of Pascal's Triangle 1 2 1. these are the coefficients of the binomial expansion of (a+b)²

= a²+2ab+b² Let a=..6 and b=1-.6=.4

then (.6)² +2(.6)(.4) = .36+.48=.84 P(at least one head given 2 flips of an unfair coin like that)=.84>.75 but<1