91Ó°ÊÓ

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

$A$and $H$are independent

$⇒P\left(A\right)=0.4$

A speaks the truth

$⇒P\left(Bâˆ\pounds A^c\right)=P\left(Hâˆ\pounds A^c\right)\stackrel{\text{ind.}}{=}P\left(H\right)=0.8$

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

$⇒P(Bâˆ\pounds A)=0.5$

The formula of total probability can be applied here (because $Aâˆ\copyright A^c=âˆ\dots ,P\left(A∪A^c\right)=1$):

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

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

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$0.68$.

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

$⇒P\left(A\right)=0.4$

A speaks the truth

$⇒P\left(Bâˆ\pounds A^c\right)=P\left(Hâˆ\pounds A^c\right)\stackrel{\text{ind.}}{=}P\left(H\right)=0.8$

$⇒P(Bâˆ\pounds A)=0.5$

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

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

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

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:

$P\left(BH∪B^c{H}^c\right)=1·P\left(A^c\right)+P\left(BH∪B^c{H}^câˆ\pounds A\right)P\left(A\right)$

$=P\left(A^c\right)+P(BHâˆ\pounds A)P\left(A\right)+P\left(B^c{H}^câˆ\pounds A\right)P\left(A\right)$

$=P\left(A^c\right)+P(Bâˆ\pounds A)P(Hâˆ\pounds A)P\left(A\right)+P\left(B^câˆ\pounds A\right)P\left(H^câˆ\pounds A\right)P\left(A\right)$

$=P\left(A^c\right)+P(Bâˆ\pounds A)P\left(H\right)P\left(A\right)+P\left(B^câˆ\pounds A\right)P\left(H^c\right)P\left(A\right)$

$=0.6+0.5·0.8·0.4+0.5·0.2·0.4$

$=0.8$

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

$P\left(Bâˆ\pounds A^c\right)=P\left(Hâˆ\pounds A^c\right)\stackrel{ind}{=}P\left(H\right)=0.8$

$⇒P(Bâˆ\pounds A)=0.5$

The definition of conditional probability:

$P(Hâˆ\pounds B)=\frac{P\left(HB\right)}{P\left(B\right)}$

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

Now again formula of total probability conditioning on $A$

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

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

$P\left(BH\right)=P(Bâˆ\pounds A){\stackrel{Áåž}{=P\left(H\right)}}^{P(Hâˆ\pounds A)}P\left(A\right)+{\stackrel{Áåž}{P\left(Hâˆ\pounds A^c\right)}}^{=P\left(H\right)}P\left(A^c\right)$

All these probabilities are known:

Substitute this into first formula:

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