Please help me with this!

Question

In a candy bag there are 50 candy cars, white, green and pink. You take two. How many candy cars of the different colors are there if the probability of taking two green cars with return is 0.25 and taking two pink is 0.09?

Answer 1

Hello, Liana,

If I understand the problem, the probability of getting two cars of the same color in a row is given for each color. That is the basis of my answer.

The probability of getting two in the row of the same color is the product of that color's percentage of the original set of 50 cars times itself. If For example if one had 100 marbles and 50 (50% or 0.50) were red, the likelihood of picking one is 50%, but the likelihood of picking two in a row is 50% x 50%, or 25% (0.25),

We are told that 2 green cars in a row has a chance of 25%, That means the original likelihood was 50% (0.5 * 0.5 = 0.25). Algebraically, this is the square root of 0.25, or 0.25^0.5. We can do the same math for the pink cars to get the original value (0.09^0.5) = 0.3 or 30%

That leaves the white cars. We know the probability of picking either a white, pink or green car is exactly 1 for our first pick. If the probabilities of that car being pink or green are 0.2 and 0.5, the white must be 0.3 (1 - .02 - .05), The three probabilities of picking a car add to one. Just multiply each probability by the total number of cars to get the number of that color. This summaries the numbers.

Correction: As per Robert Z.'s observation in the comments, below, I erred in the table The pink and white labels need to be swapped. Pink's probability is 30% and white's is 20%. The calculation in paragraph 2 "If the probabilities of that car being pink or green are 0.2 and 0.5, the white must be 0.3 (1 - .02 - .05)," should have read "If the probabilities of that car being pink or green are 0.3 and 0.5, the white must be 0.2 (1 - .03 - .05),

Sorry. And thanks to Robert for alerting everyone.

I hope this helps,

Bob