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 result 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?

$P\left(H\right)=0.8$

$A$and $H$are independent

$\Rightarrow P\left(A\right)=0.4$

A speaks the truth

$\Rightarrow P\left(B\mid A^c\right)=P\left(H\mid A^c\right)\stackrel{\text{ind.}}{=}P\left(H\right)=0.8$

$B$is independent of $H$given $A$

$\Rightarrow P(B\mid A)=0.5$

The formula of total probability can be applied here (because $A\cap A^c=\varnothing ,P\left(A\cup A^c\right)=1$):

$P\left(B\right)=P\left(B\mid A^c\right)P\left(A^c\right)+P(B\mid A)P\left(A\right)$

Formula for complement $\Rightarrow P\left(A^c\right)=1-P\left(A\right)=0.6$

Substitution of familiar probabilities:

$P\left(B\right)=0.8·0.6+0.5·0.4=0.68$.

The probability that $B$is told that the coin landed on heads is$0.68$.

$P\left(H\right)=0.8$

$\Rightarrow P\left(A\right)=0.4$

A speaks the truth

$\Rightarrow P\left(B\mid A^c\right)=P\left(H\mid A^c\right)\stackrel{\text{ind.}}{=}P\left(H\right)=0.8$

$\Rightarrow P(B\mid A)=0.5$

Again use the formula of total probability with $A$and $A^c$

$P\left(BH\cup B^c{H}^c\right)=P\left(BH\cup B^c{H}^c\mid A^c\right)P\left(A^c\right)+P\left(BH\cup B^c{H}^c\mid A\right)P\left(A\right)$

Since $A$speaks the truth if they did not forget$P\left(BH\cup B^c{H}^c\mid A^c\right)=1$.

Now use

- mutual exclusiveness of $BH$ and $B^c{H}^c$

- independence of $B$ and $H$ given $A$.

- and independence of $A$and $H$, respectively:

$P\left(BH\cup B^c{H}^c\right)=1·P\left(A^c\right)+P\left(BH\cup B^c{H}^c\mid A\right)P\left(A\right)$

$=P\left(A^c\right)+P(BH\mid A)P\left(A\right)+P\left(B^c{H}^c\mid A\right)P\left(A\right)$

$=P\left(A^c\right)+P(B\mid A)P(H\mid A)P\left(A\right)+P\left(B^c\mid A\right)P\left(H^c\mid A\right)P\left(A\right)$

$=P\left(A^c\right)+P(B\mid A)P\left(H\right)P\left(A\right)+P\left(B^c\mid A\right)P\left(H^c\right)P\left(A\right)$

$=0.6+0.5·0.8·0.4+0.5·0.2·0.4$

$=0.8$

The probability that $B$is told the correct result is $0.8$.

$P\left(B\mid A^c\right)=P\left(H\mid A^c\right)\stackrel{ind}{=}P\left(H\right)=0.8$

$\Rightarrow P(B\mid A)=0.5$

The definition of conditional probability:

$P(H\mid B)=\frac{P\left(HB\right)}{P\left(B\right)}$

From a) we know $P\left(B\right)=0.68$

Now again formula of total probability conditioning on $A$

$P\left(BH\right)=P(BH\mid A)P\left(A\right)+P\left(BH\mid A^c\right)P\left(A^c\right)$

If $A$then$B$and $H$are independent, and if $A^c$then$B\iff H$, thus:

$P\left(BH\right)=P(B\mid A){\overbrace{=P\left(H\right)}}^{P(H\mid A)}P\left(A\right)+{\overbrace{P\left(H\mid A^c\right)}}^{=P\left(H\right)}P\left(A^c\right)$

All these probabilities are known:

Substitute this into first formula:

The probability that it did in fact land on heads is $0.94$.