DEFINITIONS Two events are independent, if the probability that one event occurs in no way affects the probability of the other event occurring. Definition conditional probability: \(\displaystyle{P}{\left({A}{\mid}{B}\right)}={\frac{{{P}{\left({A}\cap{B}\right)}}}{{{P}{\left({A}\right)}}}}={\frac{{{P}{\left({A}{\quad\text{and}\quad}{B}\right)}}}{{{P}{\left({A}\right)}}}}\) SOLUTION We note that the table contains information about 8883 at-bats (given in the bottom right corner of the table). Moreover, 2110 of the 8883 at-bats are Hits, because 2110 is mentioned in the row ” Total” and in the column ” Hit” of the table. The probability is the number of favorable outcomes divided by the number of possible outcomes: \(P(Hit) = \frac{\# of\ favorable\ outcomes}{\# of\ possible\ outcomes} = \frac{2110}{8883} \approx 0.2375 = 23.75\%\)

Next, we note that 352 of the 8883 at-bats are post-season, because 352 is mentioned in the row ”Post” and in the column ”Total” of the given table. \(P(Post) = \frac{\# of\ favorable\ outcomes}{\# of\ possible\ outcomes}= \frac{352}{8883}\)

Next, we note that 87 of the 8883 at-bats are hits in post-season, because 87 is mentioned in the row ”Post” and in the column ”Hit” of the given table. \(P(Hit and Post) = \frac{\# of\ favorable\ outcomes}{\# of\ possible\ outcomes} = \frac{87}{8883}\) Use the definition of conditional probability: \(\displaystyle{P}{\left({H}{i}{t}{\mid}{P}{o}{s}{t}\right)}={\frac{{{P}{\left({H}{i}{t}{\quad\text{and}\quad}{P}{o}{s}{t}\right)}}}{{{P}{\left({P}{o}{s}{t}\right)}}}}\)

\(\displaystyle{\frac{{\frac{{87}}{{8883}}}}{{\frac{{352}}{{8883}}}}}\)

\(\displaystyle{\frac{{{87}}}{{{352}}}}\)

\(\displaystyle\approx{0.2472}\)

\(\displaystyle={24.72}\%\)

If events A and B are independent, then \(P(A|B) = P(A)\) and \(P(B|A) = P(B)\)

In this case, we note that \(P(Hit|Post) = 0.2472\) is not the same as \(P(Hit) = 0.2375\) and thus the two events are not independent.