15% of m&m's candies in a bag are yellow. If you randomly pull 3 m&ms of the bag what is the probability the first one is yellow and the next two are not?
A. 11%
B. 22%
C. 67%
D. 85%
My answer is D.
1 - 15% =0.85.
I think I'm doing it wrong
so, assume 100 candies, with 15 yellow. Your probability will be approximately
15/100 * 85/99 * 84/98 = 0.11 = 11%
Thank you
To determine the probability that the first M&M is yellow and the next two are not, we can break the problem down into steps.
Step 1: Find the probability that the first M&M is yellow:
Since 15% of the M&Ms in the bag are yellow, the probability of selecting a yellow M&M as the first one is 15% or 0.15.
Step 2: Find the probability that the next two M&Ms are not yellow:
The probability that the next M&M is not yellow can be calculated as 100% minus the probability of selecting a yellow M&M, which is 100% - 15% = 85% or 0.85. Since we need two M&Ms to be not yellow, we multiply this probability twice, resulting in (0.85 * 0.85) = 0.7225.
Step 3: Find the overall probability:
To find the overall probability, we multiply the probabilities of each step. So the overall probability is 0.15 * 0.7225 = 0.108375.
However, probabilities are typically expressed as percentages. To convert this decimal into a percentage, we multiply it by 100. So the overall probability is 0.108375 * 100 = 10.8375%.
Therefore, the correct answer is approximately 10.8%, which is closest to option A, 11%.