Question

a student must choose to participate in two different events during field day. there are four track events, two academic events, and six team sports events. what is the approximate probability that the student will choose to participate in two team sports? 0.114 0.227 0.273 0.545

Explanation:

Step1: Calculate the total number of events

The total number of events is \(4 + 2+6=12\) events.

Step2: Calculate the number of ways to choose 2 events from all events

Using the combination formula \(C(n,k)=\frac{n!}{k!(n - k)!}\), where \(n = 12\) and \(k=2\).

$$C(12,2)=\frac{12!}{2!(12 - 2)!}=\frac{12\times11}{2\times 1}=66$$

Step3: Calculate the number of ways to choose 2 team - sports events

There are \(n = 6\) team - sports events and \(k = 2\).

$$C(6,2)=\frac{6!}{2!(6 - 2)!}=\frac{6\times5}{2\times1}=15$$

Step4: Calculate the probability

The probability \(P=\frac{C(6,2)}{C(12,2)}\)

$$P=\frac{15}{66}\approx0.227$$

Answer:

0.227