91影视

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.

Let,

$F=faircoin$

$UF=unfaircoin$

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

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

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

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

$P(H/UF)=1$

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

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

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

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

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

$=\frac{1}{3}$

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

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

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.

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

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

$=\frac{1}{5}$

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

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

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

According to Baye's theorem,

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

$=1$

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