A box contains 5 coins and each has a different probability of showing heads. Le
ID: 3298630 • Letter: A
Question
A box contains 5 coins and each has a different probability of showing heads. Let p_1,... ,p_5 denote the probability of heads on each coin. Suppose that p_1 = 0, p_2 = 1/4, p_3 = 1/2, p_4 = 3/4 and p_5 = 1. Let H denote "heads is obtained" and let C_i denote the event that coin i is selected. (a) Select a coin at random and toss it. Suppose a head is obtained. What is the posterior probability that coin i was selected (i = 1,middotmiddotmiddot,5)? In other words, find P(H_2|H_1) for i = 1, ,5. (b) Toss the coin again. What is the probability of another head? In other words find P(H_2|H_1) where H_j = "heads on toss J". Now suppose that the experiment was carried out as follows: We select a coin at random and toss it until a head is obtained. (c) Find P(C_i|B_4) = "first head is obtained on toss 4."Explanation / Answer
a) First we will compute here the probability of heads to come up:
P(H) = P( H | C1)P(C1) + P( H | C2)P(C2) + P( H | C3)P(C3) + P( H | C4)P(C4) +P( H | C5)P(C5)
As the probability of choosing each of the 5 coins is equal, therefore we get:
P(C1) = P(C2) = P(C3) = P(C4) = P(C5) = 1/5 = 0.2
Also we are given the probabilities of getting head for each coin. Putting all that we get:
P(H) = 0.2*(0) + 0.2*(0.25) + 0.2*(0.5) + 0.2*(0.75) + 0.2*(1) = 0.5
The posterior probability that coin i was selected given that the head was obtained is computed as:
P(C1 | H) = P( H | C1)P(C1) / P(H) = 0
P(C2 | H) = P( H | C2)P(C2) / P(H) = 0.2*0.25 / 0.5 = 0.1
P(C3 | H) = P( H | C3)P(C3) / P(H) = 0.2*0.5 / 0.5 = 0.2
P(C4 | H) = P( H | C4)P(C4) / P(H) = 0.2*0.75 / 0.5 = 0.3
P(C5 | H) = P( H | C5)P(C5) / P(H) = 0.2*0.5 / 0.5 = 0.4
These are the required posterior probabilities for the different coins given that a head was obtained.
b) Here we are given that the same coin was tossed again, so we will have to use the posterior probabilities as the probability of getting the coin now. Therefore the probability of head again is now computed as:
P( H2 | H1) = P(C1 | H)P( H | C1) + P(C2 | H)P( H | C2) + P(C3 | H)P( H | C3) + P(C4 | H)P( H | C4) + P(C5 | H)P( H | C5)
P( H2 | H1) = 0 + 0.1*(0.25) + 0.2*(0.5) + 0.3*(0.75) + 0.4*(1) = 0.75
Therefore the required probability here is 0.75
c) Here we have to find the posterior probability of all the coins, given that there were tails in the first 3 tosses and a head on the 4th toss.
P(B4) = P( B4 | C1)P(C1) + P( B4 | C2)P(C2) + P( B4 | C3)P(C3) + P( B4 | C4)P(C4) +P( B4 | C5)P(C5)
P(B4) = 0 + 0.753*0.25*0.2 + 0.53*0.5*0.2 + 0.253*0.75*0.2 + 0 = 0.02109375 + 0.0125 + 0.00234375 = 0.0359375
Therefore now the required posterior probabilities for different coins is obtained as:
P(C1 | B4) = 0 because a head cannot be obtained even on 4th toss here.
P(C2 | B4) = P( B4 | C2)P(C2) / P(B4) = 0.753*0.25*0.2 / 0.0359375 = 0.5870
P(C3 | B4) = P( B4 | C3)P(C3) / P(B4) = 0.53*0.5*0.2 / 0.0359375 = 0.3478
P(C4 | B4) = P( B4 | C4)P(C4) / P(B4) = 0.253*0.75*0.2 / 0.0359375 = 0.06522
P(C5 | B4) = 0 because a tail cannot be obtained in this coin.
These are the required posterior probabilities for the different coins.
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