Question

Sal has a bag of hard candies: 3 are lemon and 2 are grape. He ate 2 of the candies while waiting for the bus, selecting them at random one after the another.

What is the probability of at least one candy was lemon?

Answer 1

Use complementarity.
Calculate the probability of getting <i>no lemon</i>, and subtract from 1.

Two-step experiment, second step dependent on the first:
Total number of candies at start=5
Step 1:
Number of grapes=2
Probability(No lemon)= 2/5
Step 2:
Number of candies left=4
Number of grapes left=1
Probability(no lemon)=1/4

Probability of both happening=2/5*1/4=1/10

Therefore probability of having at least one lemon = 1-1/10=9/10

Answer 2

I don't understand it says at least one lemon, how can there be none

Answer 3

There are only 3 possibilites:
- no lemon
- 1 lemon
- 2 lemon

so you want the prob (1 lemon OR 2 lemon)
= 2(3/5)(2/4) = 3/5 + (3/5)(2/4) = 3/10
= 9/10

This involved finding the prob of two cases, plus an addition.
You know that the prob of all 3 cases is 1
so what MathMate did was calculate the <b>case you don't want</b> and subtract it from 1 to also get 9/10

that involved the calculation of only one prob and a subtraction.

In this case it didn't make much difference in the length of the solution, but suppose you had 5 different cases and you wanted "at least 1 of those"
Then you would to find prob(1) + prob(2) + ... Prob (5) whereas with the "back-door" or complimentary approach you would have to find only the prob (none) and subtract that from 1.

Answer 4

Thank you