I have the following sentence:
If I toss a biased coin with 2/3 chance of landing on heads, given that there was at least one head in 3 flips, what is the probability that there is only 1 head?
If we let $X$ represent the number of heads, then I can use $Baye's$ formula to find out the probability.
Which of the following would be the formula?
$$P(X=1 | X \geq 1)$$ $$P(X \geq 1 | X=1)$$
I think its the first one, but my sheet is saying the second one. I think it might have made a mistake. Which is the right probability?
Since there is some confusion, let me post some details.
There are $8$ possible outcomes for the three tosses. As the probabilities for $T,H$ are not the same, these outcomes are not equally probable.
There are seven outcomes that have at least one $H$. They are $$HHH,HHT,HTH,THH,HTT,THT,TTH$$
The probability that we have at least one $H$ is $$P_{≥1}=1-P(TTT)=1-\left(\frac 13\right)^2=0.962962963$$
How do we compute the conditional probability? Well, in this case we simply divide the unconditional probability by $P_{≥1}$. Why? Well, in general, conditional probability is defined by $$P(A\,|\,B)=\frac {P(A\cap B)}{P(B)}$$ Here $A$ is the event "you get exactly one $H$" and $B$ is the event "you get at least one $H$". Of course, in this case $$A\cap B=A\implies P(A\cap B)=P(A)=P(HTT)+P(THT)+P(TTH)$$ Thus our answer is the ratio $$\frac {P(HTT)+P(THT)+P(TTH)}{P_{≥1}}$$
Of course all three of those events, $HTT,THT,TTH$ have the same probability, namely $\left( \frac 13 \right)^2\times \frac 23$. Thus our numerator is $3\times \left( \frac 13 \right)^2\times \frac 23=\frac 29=.222\cdots$
Finally the answer we seek is $$\frac {.2222\cdots}{0.962962963}=\boxed {0.230769231 }$$