Definition conditional probability:

\(P(B|A)=\frac{P(A\cap B)}{P(A)}=\frac{\text{P(A and B)}}{P(A)}\)

SOLUTION

\(\begin{array} {c|ccc|c} & \text { Elite } & \text { Non-elite } & \text {Did not play } & \text { Total } \\ \hline \text { Yes } & 10 & 9 & 24 & 43 \\ \text { No } & 61 & 206 & 548 & 815 \\ \hline \text { Total } & 71 & 215 & 572 & 858 \end{array}\)

We note that 10 players were elite players with arthritis, 9 players were non-elite players with arthritis and 24 players with arthritis did not play.

The probability is the number of favorable outcomes divided by the number of possible outcomes.

The probability of a player having arthritis is then the total number of players with arthritis divided by the sample size:

\(P(\text{arthritis})=\frac{\text{# of favorable outcomes}}{\text{# of possible outcomes}}\)

\(=\frac{10+9+24}{10+61+9+206+24+548}\)

\(=\frac{43}{858}\)

\(\approx0.0501\)

\(=5.01\%\)

10 of the 858 players are elite players and have arthritis. \(P(\text{ elite and arthritis})=\frac{\text{# of favorable outcomes}}{\text{# of possible outcomes}}=\frac{10}{858}\)

Use the definition of conditional probability:

\(P(\text{elite | arthritis})=\frac{\text{# of favorable outcomes}}{\text{# of possible outcomes}}\)

\(=\frac{\frac{10}{858}}{\frac{43}{858}}\)

\(=\frac{10}{43}\)

\(\approx 0.2326\)

\(=23.26\%\)