Question

A study in Sweden looked at former elite soccer players, people who had played soccer but not at the elite level, and people of the s

Two-way tables
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asked 2021-03-04

A study in Sweden looked at former elite soccer players, people who had played soccer but not at the elite level, and people of the same age who did not play soccer. Here is a two-way table that classifies these individuals by whether or not they had arthritis of the hip or knee by their mid-50s.

\(\begin{array} {c|ccc|c} & \text { Elite } & \text { Non-elite } & \text {Did not play } & \text { Total } \\ \hline \text { Yes } & 10 & 9 & 24 & 43 \\ \text { No } & 61 & 206 & 548 & 815 \\ \hline \text { Total } & 71 & 215 & 572 & 858 \end{array}\)
Suppose we choose one of these players at random. What is the probability that the player has arthritis, given that he or she was classified as an elite soccer player?

Answers (1)

2021-03-05
Definitions
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:
\(P(\text{elite | arthritis})=\frac{\text{# of favorable outcomes}}{\text{# of possible outcomes}}\)
\(=\frac{\frac{10}{858}}{\frac{43}{858}}\)
\(=\frac{10}{43}\)
\(\approx 0.2326\)
\(=23.26\%\)
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