91Ó°ÊÓ

Three players toss coins with probabilities $P_1,P_2$and $P_3$respectively for $A,B$and $C$ to get heads.

Event in question is:

And that is the probability that $A$is excluded in a single set of tosses. But we know that one of the players will be the odd man out, so this probability is the one we are looking for:

$P(A\text{is the odd man out}âˆ\pounds \text{someone is the odd man out})$

Nominator:

$P\left(A\text{is the odd man out, and someone is the odd man out}\right)$

$=P\left(A\text{is the odd man out}\right)$

$=P\left(A\text{gets different result from}B\text{and}C\right)$

$=P_1\left(1-P_2\right)\left(1-P_3\right)+\left(1-P_1\right)P_2{P}_3$

$=P_1-P_1{P}_3-P_1{P}_2+P_2{P}_3$

Denominator:

$P\left(\text{someone is the odd man out}\right)=1-P\left(\text{no one is the odd man out}\right)$

$=1-P\left(\text{all get the same result}\right)$

$=P_1{P}_2{P}_3+\left(1-P_1\right)\left(1-P_2\right)\left(1-P_3\right)$

$P(A\text{is the odd man out}âˆ\pounds \text{someone is the odd man out})$

$=\frac{P_1\left(1-P_2\right)\left(1-P_3\right)+\left(1-P_1\right)P_2{P}_3}{P_1{P}_2{P}_3+\left(1-P_1\right)\left(1-P_2\right)\left(1-P_3\right)}$

The probability of $P(A\text{is the odd man out}âˆ\pounds \text{someone is the odd man out})$

$=\frac{P_1\left(1-P_2\right)\left(1-P_3\right)+\left(1-P_1\right)P_2{P}_3}{P_1{P}_2{P}_3+\left(1-P_1\right)\left(1-P_2\right)\left(1-P_3\right)}$.