There are three coins $C_1,C_2,C_3$,$C_1$ is a fair coin painted blue on the head side and white on the tail side.$C_2$ and $C_3$ are biased coins so that the probability of a head is $p$.They are painted blue on the tail side and red on the head side.Two of the three coins are selected at random and tossed.If the probability that both the coins land up with sides of the same colour is $\frac{29}{96}$ then prove that $p=\frac{5}{8}$ or $\frac{7}{8}$.
Either $C_1,C_2$ or $C_2,C_3$ or $C_1,C_3$ coins are selected and tossed.If $C_1,C_2$ are selected,probability that both the coins land up with sides of the same colour is $\frac{1}{2}(1-p)$.
If $C_2,C_3$ are selected,probability that both the coins land up with sides of the same colour is $(1-p)^2+p^2$.
If $C_1,C_3$ are selected,probability that both the coins land up with sides of the same colour is $\frac{1}{2}(1-p)$.
So Probability that both the coins land up with sides of the same colour is $\frac{1}{2}(1-p)+\frac{1}{2}(1-p)+(1-p)^2+p^2=\frac{29}{96}$.
But this is not giving me the correct answers.What mistake have i done?
Please help me.
Best Answer
You forgot a factor $\frac13$, the probability for each of the three cases you considered. If you add that factor on the left-hand side, the answer comes out right.