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A coin having probability .8 of landing on heads is flipped. A observes the result either heads or tails and rushes off to tell B. However, with probability .4, A will have forgotten the re- sult by the time he reaches B. If A has forgotten, then, rather than admitting this to B, he is equally likely to tell B that the coin landed on heads or that it landed tails. (If he does remember, then he tells B the correct result.)

(a) What is the probability that B is told that the coin landed on heads?

(b) What is the probability that B is told the correct result?

(c) Given that B is told that the coin landed on heads, what is the probability that it did in fact land on heads?

for (a) i did something like this :

P(b is told head)=P(head)xP(remember result)+P(make a guess)xP(forget the result)

            =0.8x0.6+0.5x0.4

            =0.68

i don't know how to solve (b) and (c) part and please correct (a) weather wrong or right.

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2 Answers 2

1

Result of the flip:

  1. Heads (.8)
    A have forgotten:

    • Yes (.4)

      a) A tells "Heads" (.5) - result "Heads", true, $.8\cdot .4\cdot.5=.16$

      b) A tells "Tails" (.5) - result "Tails", false, $.8\cdot .4\cdot.5=.16$

    • No (.6) - result "Heads", true, $.8\cdot.6=.48$

  2. Tails (.2)

    • Yes (.4)

      a) A tells "Tails" (.5) - result "Tails", true, $.2\cdot .4\cdot.5=.04$

      b) A tells "Heads" (.5) - result "Heads", false, $.2\cdot .4\cdot.5=.04$

    • No (.6) - result "Tails", true, $.2\cdot.6=.12$

To obtain "Heads" we have three paths with probabilities $.16, .48, .04$, so $$P(\text{"Heads"})=.16 + .48+ .04 = .68$$

To obtaint correct result we have 4 paths with probabilities $.16,.48,.04,.16$, so $$P(\text{Correct result})=.16+.48+.04+.12=.80$$

If B was told "Heads" (with probability $.68$) we have 2 situations, that leads to "Heads" and correct result with probabilities $.16, .48$. $$P(\text{"Heads" & correct result})=.16+.48=.64$$ So $$P(\text{correct result if "Heads"})=\frac{P(\text{"Heads" & correct result})}{P(\text{"Heads"})}=\frac{.64}{.68}=\frac{16}{17}\approx .94 $$

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Since you have not shown any effort to work out the problem yourself i will give you hints for the first part.

Probability that b is told that coin lands on head is the probability that coin lands on head given that a has not forgotten and probabilty that b has reported head given that b has forgotten.

Transform all this to mathematical equation you will reach the answer.