Let R denote the event of choosing a regular coin and F denote the event of choosing the fake coin.
The prior probabilities are:
P(R)=54=0.8 P(F)=51=0.2 Let E be the event of getting two consecutive heads (HH).
The conditional probabilities are:
P(E∣R)=P(head∣R)2=0.52=0.25 P(E∣F)=P(head∣F)2=1.02=1.0 Using Bayes' theorem, the posterior probability that the chosen coin is fake given two heads is:
P(F∣E)=P(E∣F)P(F)+P(E∣R)P(R)P(E∣F)P(F) Substitute the known values:
P(F∣E)=(1.0×0.2)+(0.25×0.8)1.0×0.2=0.2+0.20.2=0.40.2=0.5 Rounded to two decimal places, the answer is 0.50 (acceptable range: 0.49 to 0.51).