ngl im super confused on which formula to use for this one if any at all since i'm so logically inept
The following M&M colors are in the bowl: 4 yellow, 6 orange, 3 green, 5 blue, 2 brown. What is the probability of selecting a brown candy, not replacing it, then selecting a blue candy?
To start, what is the probability of selecting a brown candy?
No worries! I'll help you figure out the probability step by step.
To calculate the probability of selecting a brown candy, not replacing it, and then selecting a blue candy, we need to use the concept of conditional probability.
First, let's find the probability of selecting a brown candy from the bowl. The total number of candies in the bowl is the sum of all the different colors: 4 yellow + 6 orange + 3 green + 5 blue + 2 brown = 20 candies.
The probability of selecting a brown candy as the first pick is the number of brown candies divided by the total number of candies in the bowl:
P(Brown) = Number of brown candies / Total number of candies = 2 / 20 = 1/10.
After selecting a brown candy, we move on to the second selection without replacing the candy. Now, there are 19 candies left in the bowl (since we didn't put the brown candy back).
The probability of selecting a blue candy as the second pick is the number of blue candies remaining in the bowl divided by the total number of candies left:
P(Blue|Brown) = Number of blue candies / Total number of candies left = 5 / 19.
To find the overall probability of selecting a brown candy, not replacing it, and then selecting a blue candy, we multiply the individual probabilities together:
P(Brown and Blue) = P(Brown) * P(Blue|Brown) = (1/10) * (5/19).
So the answer is P(Brown and Blue) = 1/38.
I hope this explanation helps you understand the steps to solve the problem!