Question

Problem 8. Suppose you are QC inspector for the Lucky Coin® company, who make specialized

coins for flipping. Their main product is a coin that is supposed to come up heads 75% of the
time. However, the machinery occasionally malfunctions without warning, producing a coin with
probability of heads only 50%. To attempt to sort out the good coins from the defective ones, you
implement the following QC policy: flip each coin five times and discard it if it doesn't come up
heads at least three times (otherwise pass it).
a) What is the probability of discarding good coins?
b) What is the probability of passing defective coins?
c) If your process is known to produce 20% defective coins, what fraction of coins that were
discarded are actually good?
d) If your costs are $5 for each defective coin passed and $2 for each good coin discarded, what
is the cost of this QC policy on a per-item-produced basis?
e) Relate the problem here to the discussion of diagnostic testing in the "conditioual probability
examples" notes.
() Find a cheaper QC policy (using the same flip-five-timos test), if there is one. If not, explain
why.

Answer 1

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:

`\begin{gathered} P(h\ge3)=P(h=5)+P(h=4)+P(h=3) \\ P(h\ge3)=((3)/(4))^5+5\cdot((3)/(4))^4\cdot((1)/(4))+(5!)/(3!\cdot2!)\cdot((3)/(4))^3\cdot((1)/(4))^2 \\ P(h\ge3)\approx0.9 \end{gathered}`

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:

`\begin{gathered} P(h\ge3)=P(h=5)+P(h=4)+P(h=3) \\ P(h\ge3)=((1)/(2))^5+5\cdot((1)/(2))^4\cdot((1)/(2))+(5!)/(3!\cdot2!)\cdot((1)/(2))^3\cdot((1)/(2))^2 \\ P(h\ge3)=0.5 \end{gathered}`

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:

`P(G|D)=(P(D|G)\cdot P(G))/(P(D))`

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:

`P(G|D)=(0.1\cdot0.8)/(0.18)=0.44`

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