The probability a student chose football given that they like watching sports

probabilitystatistics

This equation comes from Edgenuity's course of Statistics, and I am taking the course as a high school senior. I understand that to find $P(A$|$B)$, one divides $P(A$ and $B)$ by $P(B)$. I attempt to do this in my solving of the problem below:

Some randomly selected high school students were asked to name their favorite sport to watch. The table displays the distribution of results.

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What is the probability that a student chose football given that they like watching sports?

First, to find $P($Like watching sports and choosing football$)$, I multiplied $0.23$ and $0.84$ (found by adding all of the probabilities of those who chose a certain sport in the chart), which equals $0.1932$. I divided $0.1932$ by $P($Like watching sports$)$, which is $0.84$. $P($Choosing football | Like watching sports$)$ is then $0.23$. Are my methods and answer correct? If not, please explain why.

Best Answer

The product rule is that: the probability for the intersection of independent events equals the product of the probabilities for each event.

However, liking football and liking sport are not independent events; since if you do not like sport you certainly will not like football.   So this rule is inapplicable.

Indeed, liking football is a subset of liking sport.   So the event of doing both is simply the event of liking football. That has a probability of $0.23$.

So by definition of conditional probability; $$\mathsf P(\text{Like(Football)}\mid\text{Like(A Sport)})=\dfrac{23}{84}$$