a) Good coins are supposed to come up heads 3/4 = 75% of the time.
Them, the probability of passing good coins is given by:
[tex]\begin{gathered} P(h\ge3)=P(h=5)+P(h=4)+P(h=3) \\ P(h\ge3)=(\frac{3}{4})^5+5\cdot(\frac{3}{4})^4\cdot(\frac{1}{4})+\frac{5!}{3!\cdot2!}\cdot(\frac{3}{4})^3\cdot(\frac{1}{4})^2 \\ P(h\ge3)\approx0.9 \end{gathered}[/tex]
Then, the probability of discarding good coins is 1 - 0.9 = 0.1
b) Defective coins are supposed to come up heads 1/2 = 50% of the time.
Them, the probability of passing good coins is given by:
[tex]\begin{gathered} P(h\ge3)=P(h=5)+P(h=4)+P(h=3) \\ P(h\ge3)=(\frac{1}{2})^5+5\cdot(\frac{1}{2})^4\cdot(\frac{1}{2})+\frac{5!}{3!\cdot2!}\cdot(\frac{1}{2})^3\cdot(\frac{1}{2})^2 \\ P(h\ge3)=0.5 \end{gathered}[/tex]
c) We know that 20% of the coins are defective.
To finde the fraction of coins that were discarded, we can aply the bayes theorem. Let P(G|D) be the probability of getting a good coin given it was discarded. Then we have:
[tex]P(G|D)=\frac{P(D|G)\cdot P(G)}{P(D)}[/tex]
From part a, we know that P(D|G) = 0.1. And we also know that 80% of the coins are good, then P(G) = 0.8. Since 20% of the coins are defective, we have P(D) = P(G and D)+P(Defective and D) = 0.8*0.1 + 0.2*0.5 = 0.18
Them we have:
[tex]P(G|D)=\frac{0.1\cdot0.8}{0.18}=0.44[/tex]
d) The cost is P(G)*P(D|G)*2 + P(Defective)*(1-P(D|Defective))*5 = 0.8*0.1*2 + 0.2*0.5*5 = $0.66
