Answer
The probability of choosing an athlete that plays either Basketball or Football is:
[tex]P(A\text{ OR B) = }\frac{33}{50}=66\text{ \%}[/tex]SOLUTION
Problem Statement
The question tells us that a survey revealed that 63% of athletes are footballers, 34% are basketball players while 31% can play both sports.
We are required to find the probability that a randomly chosen athlete can play either basketball or football.
Method
The probabilities of playing basketball and football have been given. Let us take these as two separate events A and B.
We are being asked to find the probability of playing basketball or football; that is, the probability of getting event A or event B.
This question is clearly asking for the OR probability of events A and B, which is given by the formula below:
[tex]P(A\text{ OR B) = P(A) + P(B) - }P(A\cap B)[/tex]Implementation
To solve the question, we just simply use the formula above. This is done as follows:
1 . First, we shall list out our variables:
[tex]\begin{gathered} \text{If A = Event that the chosen athlete is a footballer} \\ B=Event\text{ that the chosen athlete plays basketball} \\ \\ P(A)=63\text{ \% = }\frac{63}{100} \\ \\ P(B)=34\text{ \% = }\frac{34}{100} \\ \\ P(A\cap B)=31\text{ \% = }\frac{31}{100} \end{gathered}[/tex]2. Apply the formula:
[tex]\begin{gathered} P(A\text{ OR B) = P(A) + P(B) - }P(A\cap B) \\ P(A\text{ OR B) = }\frac{63}{100}+\frac{34}{100}-\frac{31}{100}=\frac{63+34-31}{100} \\ \\ \therefore P(A\text{ OR B) = }\frac{66}{100}=\frac{33}{50} \end{gathered}[/tex]Final Answer
The probability of choosing an athlete that plays either Basketball or Football is:
[tex]P(A\text{ OR B) = }\frac{33}{50}=66\text{ \%}[/tex]