A bag of M&Ms contains 12 red, 11 yellow, 5 green, 6 orange, 5 blue, and 16 brown candies. What is the probability that if you choose 2 M&Ms from the bag (one after the other) without looking, you will choose 2 yellow ones?
First calculate the total numebr of candies:
12+11+5+6+5+16=55
(check my work, because my calculator sometimes misses a digit).
There are 11 yellow candies out of the 55.
If you pick (not choose) without looking one after another, this is what is called pick without replacement.
If two yellow candies are required, the probability to pick the first one is therefore 11/55.
Since there are now 10 left out of 54 (without replacement), the probability of picking a yellow this time is 10/54.
The probability that both events will occur is therefore the product of the two, namely
11/55*10/54.
I will leave you to calculate the final answer. Don't forget to cancel common factors if you present it in fraction form (my preference).
To find the probability of choosing 2 yellow M&Ms from the bag, we need to determine the total number of possible outcomes and the number of favorable outcomes.
Step 1: Find the total number of possible outcomes:
When choosing two M&Ms without looking, the first selection can be any of the 49 candies in the bag. After the first choice, there will be 48 candies remaining for the second selection.
Therefore, the total number of possible outcomes is 49 * 48 = 2352.
Step 2: Find the number of favorable outcomes:
Since we want to choose 2 yellow M&Ms, the first selection can be any of the 11 yellow candies, and for the second selection, there will be 10 yellow candies remaining.
Therefore, the number of favorable outcomes is 11 * 10 = 110.
Step 3: Calculate the probability:
Probability = Number of favorable outcomes / Total number of possible outcomes.
Plugging in the values, we get:
Probability = 110 / 2352 = 0.0467.
So, the probability of choosing 2 yellow M&Ms from the bag is approximately 0.0467 or 4.67%.