# Thread: Having trouble with this probability question...

1. ## Having trouble with this probability question...

A poll found that the probability that a student chosen at random had visited the arcade within the past week was 0.3. The probability that a student chosen at random played games on a home-based game machine was 0.5. The probability that a student did both was 0.2. What was the probability that a student played on an arcade machine or on a home-based machine?

Thanks in advance!

2. I believe the probability of either event A or B occurring is

$P(A) + P(B) - P(A + B)$,

which in this case is

$0.3 + 0.5 - 0.2 = 0.6$.

3. Originally Posted by PiratePrincess
A poll found that the probability that a student chosen at random had visited the arcade within the past week was 0.3. The probability that a student chosen at random played games on a home-based game machine was 0.5. The probability that a student did both was 0.2. What was the probability that a student played on an arcade machine or on a home-based machine?

Thanks in advance!
Assuming that a student who visits the arcade always plays on an arcade machine, then

0.3=P(plays on arcade machine)=P(plays only on arcade machine)+P(plays on arcade and home machine)

0.5=P(plays on home machine)=P(plays at home only)+P(plays at home and at arcade)

0.3+0.5=P(arcade only)+P(home only)+2P(both)

0.8=P(plays either one)+P(both)

P(plays either one)=0.8-0.2