Probability of At Least 1 Cherry Yogurt: 7/55

[Gringo123]

A box contains 2 strawberry yogurts, 4 vanilla yogurts and 6 cherry yogurts.
3 yogurts are selected at random form the box. (they are not replaced)
Calculate the probability that at least 1 of the selected yogurts is a cherry yogurt.

This is how I have approached the problem. Did I get it right?

Pr. that all 3 are cherry = 6/12 x 5/11 x 4/10 = 1/11

Pr. that only 2 are cherry = 6/12 x 5/11 x 6/10 = 3/22 ( I have calculated that the P would 3/22 for all possible combinations - i.e. 1st 2 selected, last 2 selected, 1st and 3rd)

Pr. that only 1 is cherry = 6/12 x 6/11 x 5/10 = 3/22 (again, all combinations = same probability)

1/11 + 3/22 + 3 /22 = 7/55

 

[jbunniii]

This way is perhaps easier:

P(at least one is a cherry yogurt) = 1 - P(none is a cherry yogurt)

P(none is a cherry yogurt) = 6/12 x 5/11 x 4/10 = 1/11

Thus

P(at least one is a cherry yogurt) = 1 - 1/11 = 10/11

Since this does not match your answer, we can conclude that at least one of your probabilities is incorrect.

Your P(all 3 are cherry) is fine, so the error is with one (or both) of the other two.

 

[jbunniii]

OK, I see the problem.

P(exactly 1 is a cherry yogurt) =

P(1st is cherry, 2nd is not, 3rd is not) +
P(1st is not, 2nd is cherry, 3rd is not) +
P(1st is not, 2nd is not, 3rd is cherry)

EACH of these terms has probability 3/22 (as you calculated).

You have a similar error with P(exactly 2 are cherry yogurts).