Question

of 50 students going on a class trip, 35 are student - athletes and 5 are left - handed. of the student athletes, one of the students on the trip is an athlete or is left - handed? athletes left - handed 0.2 0.74 0.5 0.8

Explanation:

Step1: Recall probability formula

The formula for $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. Here, $n = 50$, $n(A)=35$, $n(B)=5$. Assume $n(A\cap B) = 2$ (since no information about intersection is given, we assume some non - negative value; worst - case non - overlapping would be wrong as we need to use the inclusion - exclusion principle. A reasonable small non - zero value is assumed for illustration. In a real - world scenario, more data might be needed. But if we assume no overlap between athletes and left - handers, we can still calculate).
$P(A)=\frac{35}{50}$, $P(B)=\frac{5}{50}$, $P(A\cap B) = 0$ (assuming non - overlapping for simplicity as no intersection data).

Step2: Calculate $P(A\cup B)$

$P(A\cup B)=\frac{35}{50}+\frac{5}{50}-0=\frac{35 + 5}{50}=\frac{40}{50}=0.8$

Answer:

$0.8$