Question

6. Jar of Coins. 100 quarters. Most of Coins. Note: Thiadaned from a Google interview question. I have a jar Most of them are ar fi luarters, but 12 of them are takel of these fala eads on both sides and 5 have tails on both sides. quarters, 7 have heads on both sides and 5 have a) If you select a coin from the cct a coin from the jar at random and flip it 3 times and get heads every time, at is the probability that this is a two-heads" quarter b) Repeat the calculation in mart al hut this time you flip the coin times, getting heads each time. What is the probability that this is a two-heads" quarter (in terms of k)? c) Using a computer, plot this probability as a function of k. For what value of k is the probability more than 1/2? For what value of k is the probability more than 0.92

Answer 1

a,b. We solve the problem for k and plug in k = 3 to get the answer to the 1st part. Let All be the event in which we flip a coin k times and get a head each time. Let F be the event that we draw a "two heads" quarter. We are interested in finding the probability that the event F has oc- cured given that Ax has occured. We use Bayes rule to compute this probability: P[F|A) = P[A |F] * P[F] ? P[4x] 5 1 88 0 * 100+ 24 100 Thus, the probability that the coin is a two heads quarter is 7 Plugging in k =3, we get the answer to part a) as — c. Note that the probability is increasing in k. From the graph below, we can see that for k > 4, P[FA] > 0.5 and for k > 7, P[FA] > 0.9.

p 0.2 0.4 0.6 0.8 1.0 ο ο ο ο ο Ι ο ο Ο 10 20 30 000000000000000000000000000οοοοοοοοοοοοοο 40

The R code for the graph

p=vector()
k=seq(1,50,1)
for( i in 1:50)
{
p[i]=7*2^(i)/(7*2^(i)+88)
}

plot(k, p)
abline(h=0.5)
abline(h=0.9)