(PRACTICE Problem) Section 3.5 & 3.6, Problem 5: Suppose that a random experimen

Question

(PRACTICE Problem) Section 3.5 & 3.6, Problem 5: Suppose that a random experiment consists of randomly selecting one of four coins C1. C2. C3. and C4. tossing it and observing whether a head (H) or a tail occurs. Further suppose that the coins C1. C2. and C3 are biased such that the probabilities of a head occurring for coins C1. C2. and C3 e 0.90, 0.75, 0.60, respectively while the fourth coin C4 is a fair coin. (a) If the outcome of the experiment was a head, find the probability that coin C2 was tossed (b) If the outcome of the experiment was a tail, find the probability that coin C4 was tossed First, what are the following marginal probabilities P(C1) = P(C2) PC3,- Second, what are the following conditional s (read them carefully): PH , I C2,- PlH'IC4) = Now, what is the marginal probability of a head, P(H) and what is the marginal probability of a tail Finally, what are the requested results? (a) If the outcome of the experiment was a head, find the probability that coin C2 was tossed, P(C2 I H)- (b) If the outcome of the experiment was a tail, find the probability that coin C4 was tossed, PIC4 IH) , P(H)

Explanation / Answer

Here we are given that:

P( H | C1) = 0.9, P( H | C2) = 0.75 and P(H | C3) = 0.6.

Als as C4 coin is fair, therefore P( H | C4) = 0.5

Now using law of total probability, we get:

P(H) = P( H | C1) P(C1) + P( H | C2) P(C2) + P( H | C3) P(C3) + P( H | C4) P(C4)

As each of the coin is randomly picked, therefore P(C1) = P(C2) = P(C3) = P(C4) = 0.25

Therefore, we get:

P(H) = 0.25*(0.9 + 0.75 + 0.6 + 0.5) = 0.6875

a) Now given that heads comes up, probability that the coin C2 was tossed is computed using Bayes theorem as:

P( C2 | H) = P( H | C2) P(C2) / P(H) = 0.75*0.25 / 0.6875 = 0.2727

Therefore 0.2727 is the required probability here.

b) Now We know already that P(H) = 0.6875, therefore P(T) = 0.3125

Now using Bayes theorem, we get:

P( C4 | T) = P(T | C4) P(C4) / P(T) = 0.5*0.25 / 0.3125 = 0.4

Therefore 0.4 is the required probability here.

For the next few marginal probabilities we have already found above as:

P(C1) = P(C2) = P(C3) = P(C4) = 0.25

P( H | C1) = 0.9, P( H | C2) = 0.75, P(H | C3) = 0.6, P( H | C4) = 0.5

Also we have already found that: P(H) = 0.6875 and therefore P(H') = 0.3125

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