A certain store sells small, medium and large toy trucks in each of the colors red blue green and yellow. The store has an equal number of trucks of each possible color-size combination. If Paul wants a medium red truck and his mother will randomly select one of the trucks in the store, what is the probability that the truck she selects will have at least one of the two features Paul wants?
A) 1/4
B) 1/3
C) 1/2
D) 7/12
E) 2/3
Hi! This question is a great illustrator of the "one minus" approach to complicated probability problems.
When you see "at least" or "at most" in probability, you're going to be solving for multiple scenarios. However, it's often quicker to solve for what you do NOT want to happen and then subtract from 1, since:
Prob(want) + Prob(don't want) = 1
and
Prob(want) = 1 - Prob(don't want)
In this case we want the truck to have at least one of two features, so what we do NOT want is a truck with neither feature.
Prob(not red) = 3/4 and Prob(not medium) = 2/3
Since we want (not red) AND (not medium) we MULTIPLY to get:
3/4 * 2/3 = 6/12 = 1/2
So, the probability of getting at least one of those two features is:
1 - 1/2 = 1/2
choose (C)!
I'm curious as to the source for this question, since it's kinder and gentler than most - usually a trap is to forget to do the final step (subtract from 1), but since the answer is 1/2 you don't get punished for forgetting!
As some good probability practice, let's calculate the long way as well:
Prob(at least 1 of red/medium) = prob(red not medium) + prob(medium not red) + prob(red AND medium)
= (1/4 * 2/3) + (3/4 * 1/3) + (1/4 * 1/3)
= 2/12 + 3/12 + 1/12
= 6/12
= 1/2
As you can see, the "one minus" approach is a lot less work!
Stuart
