What is the probability of right and left handedness given these probabilities?

Question

Me and some friends were having a debate over one of the questions presented during a lecture. According to my one of friend, who attended the lecture, he said that the professor said the correct answer was $\frac{1}{16}$ and that $\frac{1}{4}$ was a "trick" answer.
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The way my friends approached the problem was to sketch out all the sample space, where L = Left-handed and R = Right-handed. $$RR =\frac{9}{16}$$ $$RL =\frac{3}{16}$$ $$LR =\frac{3}{16}$$ $$LL =\frac{1}{16}$$ Then due to them knowing that $RR$ will never be an option, they removed it from the list, updated the denominator to 7 and got: $$LL = \frac{1}{7}$$ Claiming the professor got none of the correct answers on his slide.

My approach was a little different. I saw it as a conditional probability problem. So I used Bayes Theorem: $$P(A|B) = \frac{P(B|A)P(A)}{P(B)}$$ and ended out with: $$LL = \frac{1}{16}$$
If anyone could please clarify on some of the theory behind this question and how you arrived at your result, what events you created etc, it would be a great help. We have really been boggled by this question, since both of our answers made sense to us.
Thank you in advance.

Answer 1

OK, let's do this as a conditional probability question. The events in which we are interested are:

$B,$ at least one of the team's pitchers is left-handed. (That's a direct quote from the slide in the lecture.)

$A,$ the team has two left-handed pitchers. (That's what happens if you know at least one pitcher is left-handed and then you find out the other also is left-handed.)

We then have probabilities $$ P(A) = \frac{1}{16} $$ (assuming each pitcher has an independent $\frac14$ chance to be left-handed), $$ P(B) = 1 - P(\text{both are right-handed}) = 1 - \left(\frac34\right)^2 = \frac{7}{16},$$ and $$ P(B\mid A) = 1 $$ since if $A$ is true then both pitchers are left-handed and it is then certainly true that at least one of them is left-handed. Therefore $$ P(A\mid B) = \frac{P(B\mid A)P(A)}{P(B)} = \frac{1\times\frac{1}{16}}{\frac{7}{16}} = \frac17,$$ so your friend is correct and the slide in the lecture is wrong.

Answer 2

$\newcommand{\Prob}{\mathbb{P}}$You should end up getting $1/7$ using your Bayes theorem method. If $A$ is the event of having two left-handers ($LL$) and $B$ is the event that there is at least one left-hander, then $\Prob(A) = 1/16, \Prob(B) = 1 - \Prob(RR) = 1-9/16 = 7/16$, and $\Prob\left(B\mid A\right) = 1$.