Question

(a) A gambler has a fair coin and a two-headed coin in his pocket. He selects one of the coins at random; when he flips it, it shows heads. What is the probability that it is the fair coin?

(b) Suppose that he flips the same coin a second time and, again, it shows heads. Now what is the probability that it is the fair coin?

(c) Suppose that he flips the same coin a third time and it shows tails. Now what is the probability that it is the fair coin?

Short answer

(a) The probability that it is the fair coin will be $\frac{1}{3}$

(b) The probability that it is the fair coin when he flips the same coin a second time and, again, it shows heads will be $\frac{1}{5}$.

(c) The probability that it is the fair coin when he flips the same coin a third time and it shows tails is$1$.

Step-by-step solution

Given information (part a)

A gambler has a fair coin and a two-headed coin in his pocket. He selects one of the coins at random; when he flips it, it shows heads.

Solution (Part a)

Let,

$F=faircoin$

$UF=unfaircoin$

$P\left(F\right)=\frac{1}{2}$$P\left(H\right)=\frac{1}{2}$

$P\left(UF\right)=\frac{1}{2}$

$P(H/F)=\frac{P\left(F\right)\times P\left(H\right)}{P\left(F\right)}$

$\frac{\left(\frac{1}{2}\right)\times \left(\frac{1}{2}\right)}{\left(\frac{1}{2}\right)}=\frac{1}{2}$

$P(H/UF)=1$

Final solution (Part a)

Since it is an unjust coin, it includes both sides are leads.

So we need to find $\mathrm{P}(\mathrm{F}/\mathrm{H})$

$P(F/H)=\frac{P\left(F\right)\times P(H/F)}{P\left(F\right)\times P(H/F)+P\left(UF\right)\times P(H/UF)}$

$=\frac{\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)}{\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)\left(1\right)}$

$=\frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{2}}$

$=\frac{1}{3}$

Final answer (Part a)

The probability that it is the fair coin will be$\frac{1}{3}$.

Given in formation (Part b)

Suppose that he flips the same coin a second time and, again, it shows heads.

Solution (part b)

We need to find$\mathrm{P}(\mathrm{F}/\mathrm{H}/\mathrm{H})$$\mathrm{P}(\mathrm{F}/\mathrm{H}/\mathrm{H})$

By using the Baye's theorem.

$P(F/H/H)=\frac{P\left(F\right)\times P(H\cap H)}{P\left(F\right)\times P(H\cap H)+P\left(UF\right)P(H\cap H)}$

$=\frac{\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)}{\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)\left(1\right)}$

$=\frac{1}{5}$

Final answer (Part b)

The probability that it is the fair coin when he flips the same coin a second time and, again, it shows heads will be$\frac{1}{5}$.

Given information (Part c)

Suppose that he flips the same coin a third time and it shows tails.

Solution (Part c)

We need to find $\mathrm{P}(\mathrm{F}/\mathrm{T})$

According to Baye's theorem,

^{$P(F/T)=\frac{P\left(F\right)\times P(T/F)}{P\left(F\right)\times P(T/F)+P\left(UF\right)\times P(T/UF)}$}

$=\frac{P\left(F\right)\times P(T/F)}{P\left(F\right)P(T/F)+0}$

$=1$

Final answer (Part c)

The probability that it is the fair coin when he flips the same coin a third time and it shows tails is$1$.