Question

Suppose a girl throws a die. if she gets a 5 or 6, she tosses a coin 3 times and notes the number of heads if she gets 1, 2, 3, or 4 she tosses a coin once and notes whether a head or tail is obtained. if she obtained exactly one head, what is the probability that she threw 1, 2, 3, or 4 with the die

Answer 1

Let E1E1 be the event that the girl gets 5 or 6 on the roll. P(E1E1) = 26=1326=13
Let E2E2 be the event that the girl gets 1,2,3 or 4 on the roll. P(E1E1) = 46=2346=23
Let A be the event that she obtained exactly one head.
If she tossed a coin 3 times and exactly 1 head showed up, then the total number of favourable outcomes = {(HTT), (THT), (TTH),} = 3.
P (A|E1E1) = 3838 (Note: in 3 coin tosses, total sample space = 8)
If she tossed the coin only once and exactly 1 showed up, the total number of favourable outcomes = 1.
P (A|E2E2) = 1212
We need to find the probability that she threw 1, 2, 3 or 4 with the die, given that she got exactly one head.
We can use Baye's theorem, according to which P(E2|A)=P(E2)(P(A|E2)P(E1)P(A|E1)+P(E2)P(A|E2)P(E2|A)=P(E2)(P(A|E2)P(E1)P(A|E1)+P(E2)P(A|E2)
P(E2E2|A) = 12.2312.23+38.1312.2312.23+38.13 = 1313+181313+18 = 811