Jar O' Coins

We are told that the coin flipped gave 10 heads in a row. P(10 heads)= $\frac { 999 }{ 1000 } \times { \left( \frac { 1 }{ 2 } \right) }^{ 10 }+\frac { 1 }{ 1000 }$

From this, we can find out the following:

P(fair coin) = $\frac { \frac { 999 }{ 1000 } \times { \left( \frac { 1 }{ 2 } \right) }^{ 10 } }{ \frac { 999 }{ 1000 } \times { \left( \frac { 1 }{ 2 } \right) }^{ 10 }+\frac { 1 }{ 1000 } } =\frac { 999 }{ 2023 }$

P(double headed) = $\frac { \frac { 1 }{ 1000 } }{ \frac { 999 }{ 1000 } \times { \left( \frac { 1 }{ 2 } \right) }^{ 10 }+\frac { 1 }{ 1000 } } =\frac { 1024 }{ 2023 }$

Knowing this, we can then find out the probability of the next toss of the coin being a head: P(head) = $\frac { 999 }{ 2023 } \times \frac { 1 }{ 2 } +\frac { 1024 }{ 2023 } =\frac { 3047 }{ 4046 }$

Let A denote the event that the unbiased coin is chosen. B denote the event that first 10 tosses results in head, and C denote the event that 11th toss results in head. Now, we need to find P(C) observe that $P(C)=P(C|A)P(A)+P(C|A^c)P(A^c)$ but since it is given that the event B has occurred, we update P(A) with P(A/B) and $P(A^c)$ with $P(A^c|B)$ Thus, $P(C)=P(C|A)P(A|B)+P(C|A^{ c })P(A^{ c }|B$ Now, we have $P(A)=\frac { 1 }{ 1000 } ;\quad P(A^c)=\frac{999}{1000}\\P(B|A)=1\quad\quad P(B|A^c)=\frac{1}{2^{10}}=\frac{1}{1024}\\$ using bayes theorem $P(A|B)=\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|A^c)P(A^c)}\\P(A^c|B)=\frac{P(B|A^c)P(A^c)}{P(B|A)P(A)+P(B|A^c)P(A^c)}\\$ A little arithmetic gives $P(A|B)=\frac { 1024 }{ 2023 } \\ P(A^{ c }|B)=\frac { 999 }{ 2023 }\\$ using these values on the 2nd equation, and after little more computations we get $P(C)=\frac{3047}{4046}=0.753\\$