Question

You have
placed three coins in a
jar: one normal, one double-headed
and one double tailed .

someone selects a
coin
uniformly at
random from the jar and tosses it.
If it comes up heads, what's the
probability the coin is fair?​

Answer 1

Given Question :-

You have placed three coins in a jar: one normal, one double-headed and one double tailed . Someone selects a

coin uniformly at random from the jar and tosses it. If it comes up heads, what's the probability the coin is fair?

Solution :-

Let,

$\rm :\longmapsto\:E_1 : coin \: selected \: is \: normal \: coin$

$\rm :\longmapsto\:E_2 : coin \: selected \: is \: double \: headed \: coin$

$\rm :\longmapsto\:E_3 : coin \: selected \: is \: double \: tailed \: coin$

$\rm :\longmapsto\:A : getting \: head \: when \: coin \: is \: tossed.$

Thus,

$\longmapsto \: \sf P(E_1)= \: \dfrac{1}{3}$

$\longmapsto \: \sf P(E_2)= \: \dfrac{1}{3}$

$\longmapsto \: \sf P(E_3)= \: \dfrac{1}{3}$

Now,

Probability of getting head when normal coin is tossed,

$\longmapsto \: \sf P(A | E_1) = \dfrac{1}{2}$

Probability of getting head when double headed coin is tossed,

$\longmapsto \: \sf P(A | E_2) = \dfrac{2}{2} = 1$

Probability of getting head when double tailed coin is tossed,

$\longmapsto \: \sf P(A | E_3) = \dfrac{0}{2} = 0$

Now, by using Baye's Theorem, we have

$\sf \: P(E_1 | A) = \dfrac{P(E_1) . P(A | E_1)}{P(E_1) . P(A | E_1) + P(E_2) . P(A | E_2) + P(E_3) . P(A | E_3)}$

$\rm :\longmapsto\:\: P(E_1 | A) = \dfrac{\dfrac{1}{3} \times \dfrac{1}{2} }{\dfrac{1}{3} \times \dfrac{1}{2} + \dfrac{1}{3} \times 1 + \dfrac{1}{3} \times 0}$

$\rm :\longmapsto\:\: P(E_1 | A) = \dfrac{\dfrac{1}{6} }{\dfrac{1}{6} + \dfrac{1}{3}}$

$\rm :\longmapsto\:\: P(E_1 | A) = \dfrac{\dfrac{1}{6} }{\dfrac{1 + 2}{6}}$

$\rm :\longmapsto\:\: P(E_1 | A) = \dfrac{\dfrac{1}{6} }{\dfrac{3}{6}}$

$\bf\implies \:\:\: P(E_1 | A) = \dfrac{1}{3}$