Answer:
a) [tex]\frac{1}{3}[/tex] ≈0.33
b) [tex]\frac{1}{5}[/tex] = 0.2
c) 1
Step-by-step explanation:
This is a conditional probability problem.
Let P(F)= the probability that the gambler selects fair coin
P(TH) = the probability that the gambler selects two head coin
P(H|TH)= the probability of getting heads given that the gambler selects two headed coin.
P(H|F)= probability of getting heads given that the gambler selects fair coin
P(F|H)= probability of the gambler selects fair coin given that it shows head after flipping
P(H)= the probability of getting heads
P(HH)=the probability of getting two heads
Since gambler selects coins randomly P(F)=P(TH)= 0.5, P(H|F)=0.5 since the coin is fair.
We have also this conditional probability equation:
P(F|H) = [tex]\frac{P(F)*P(H|F)}{P(H)}[/tex]
a) For the first case P(H)= P(H|F)×P(F)+P(H|TH)×P(TH)=[tex]\frac{1}{2}[/tex]×[tex]\frac{1}{2}[/tex]+1×[tex]\frac{1}{2}[/tex]=[tex]\frac{3}{4}[/tex] =0.75
If we put the numbers in conditional probability formula:
P(F|H) = [tex]\frac{0.5*0.5}{0.75)}[/tex] = [tex]\frac{1}{3}[/tex]≈0.33
b) For flipping twice:
P(HH)= P(HH|F)×P(F)+P(HH|TH)×P(TH) = [tex]\frac{1}{4}[/tex]×[tex]\frac{1}{2}[/tex]+1×[tex]\frac{1}{2}[/tex]=[tex]\frac{5}{8}[/tex] =0.625
then our formula becomes:
P(F|HH) = [tex]\frac{0.5*0.25}{0.625)}[/tex] =[tex]\frac{1}{5}[/tex] =0.2
c) For flipping third time:
When flipping for the third time and it shows tails, then the coin cannot be two heads coin since it has 0 probability of getting tails. Therefore for this case the P(F|T)= the probability of choosing fair coin given that it shows tails)=1
