Solution Found!
Die A has 4 red and 2 white faces, whereas die B has 2 red
Chapter 3, Problem 83P(choose chapter or problem)
Problem 83P
Die A has 4 red and 2 white faces, whereas die B has 2 red and 4 white faces. A fair coin is flipped once. If it lands on heads, the game continues with die A; if it lands on tails, then die B is to be used.
(a) Show that the probability of red at any throw is .
(b) If the first two throws result in red, what is the probability of red at the third throw?
(c) If red turns up at the first two throws, what is the probability that it is die A that is being used?
Questions & Answers
QUESTION:
Problem 83P
Die A has 4 red and 2 white faces, whereas die B has 2 red and 4 white faces. A fair coin is flipped once. If it lands on heads, the game continues with die A; if it lands on tails, then die B is to be used.
(a) Show that the probability of red at any throw is .
(b) If the first two throws result in red, what is the probability of red at the third throw?
(c) If red turns up at the first two throws, what is the probability that it is die A that is being used?
ANSWER:
Solution
Step 1 of 3
Given that the die A has 4 red and 2 white faces
And the die B has 2 red and 4 white faces
The game continue with the die A if the coin lands on heads
The game continue with the die B if the coin lands on tails
a) we have to show that the probability of red at any through is ½
Probability of getting head and tail is same that is 1/2
P(red)=P(head)P(red if A is used)+P(tail) P( red if B is used)
=(1/2)(4/6)+(1/2)(2/6)
=1/2
Hence we proved that the probability of red at any through is 1/2