A bag contains 12 candies of which are blue . Two candies are chosen from the bag without replacement. What is the probability
1) both are blue
2) atleast one is blue
Duh! How many are blue?
both are blue
My mistake 3 are blue
To find the probability in this scenario, we need to first determine the total number of possible outcomes and the number of favorable outcomes.
1) To find the probability that both candies are blue, we need to calculate the number of ways we can choose two blue candies from a bag of 12 candies.
The total number of candies in the bag is 12. We need to choose 2 blue candies, so there are 12C2 ways to choose 2 candies out of 12. Here, 12C2 means the combination of 12 items taken 2 at a time.
The number of favorable outcomes (choosing 2 blue candies) is represented by 2C2, which means choosing 2 blue candies out of the 2 available blue candies.
Therefore, the probability that both candies chosen are blue is:
P(both are blue) = 2C2 / 12C2
2) To find the probability that at least one candy is blue, we need to calculate the number of ways we can choose at least one blue candy from a bag of 12 candies.
The total number of candies in the bag is still 12. We want to choose at least one blue candy, so we need to consider two scenarios:
- Choosing only one blue candy: 2C1 ways to choose 1 blue candy out of 2 available blue candies.
- Choosing both blue candies: 2C2 ways to choose 2 blue candies out of 2 available blue candies.
The number of favorable outcomes (choosing at least one blue candy) is the sum of the number of ways to choose only one blue candy and the number of ways to choose both blue candies.
Therefore, the probability of choosing at least one blue candy is:
P(at least one is blue) = (2C1 + 2C2) / 12C2