Q.3.14 - A First Course in Probability

Question

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? 

Step-by-Step Solution

Answer

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

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

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

Step 1: Given Information (Part a)

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

$A$ 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 $A$

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

Step 2: Explanation (Part a)

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:

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

Step 3: Final Answer (Part a)

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

Step 4: Given Information (Part b)

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

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

A speaks the truth

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

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

Step 5: Explanation (Part b)

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:

Step 6: Explanation (Part b)

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

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

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

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

$=0.6+0.5·0.8·0.4+0.5·0.2·0.4$

$=0.8$

Step 7: Final Answer (Part b)

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

Step 8: Given Information (Part c)

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

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

Step 9: Explanation (Part c)

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:

$P\left(BH\right)=0.5·0.8·0.4+0.8·0.6=0.64$

Substitute this into first formula:

$P(H\mid B)=\frac{P\left(HB\right)}{P\left(B\right)}=\frac{0.64}{0.68}=\frac{16}{17}$

Step 10: Final Answer (Part c)

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