a student must choose to participate in two different events during field day. there are four track events…

Answer:

B. 0.227

Explanation:

Step1: Calculate total number of events

$4 + 2+6=12$

Step2: Calculate number of ways to choose 2 events out of total

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

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

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

Step4: Calculate probability

$P=\frac{C(6, 2)}{C(12, 2)}=\frac{15}{66}\approx0.227$