A bag contains 8 blue and 2 red balls. Three balls are chosen at the same time at random from the bag. What is the probability that exactly two of the balls are the same colour?
you can basically get two of each color 3 diff ways
2 red = (2/10 * 1/9 * 8/8)+(8/10 * 2/9 * 1/8)+(2/10 * 8/9 * 1/8) = 3(16/720)
2 blue = (8/10 * 7/9 * 2/8)+(8/10 * 2/9 * 7/8)+(2/10 * 8/9 * 7/8) = 3(112/720)
at the end you jsut add these together and get the final %
3(16/720) + 3(112/720) = 8/15
Ah, the joys of probability! So, let's break it down.
To determine the probability of exactly two balls being the same color, we need to consider two scenarios:
Scenario 1: Two blue balls and one red ball.
The probability of choosing two blue balls out of the eight blue balls is (8/10) * (7/9). And the probability of choosing one red ball out of the two red balls is (2/8). So, multiplying these probabilities together, we get (8/10) * (7/9) * (2/8).
Scenario 2: Two red balls and one blue ball.
Similarly, the probability of choosing two red balls out of the two red balls is (2/10) * (1/9). And the probability of choosing one blue ball out of the eight blue balls is (8/8). Multiplying these probabilities together, we get (2/10) * (1/9) * (8/8).
Now, all we need to do is add the probabilities of these two scenarios together to get the final probability.
Therefore, the probability of exactly two balls being the same color is (8/10) * (7/9) * (2/8) + (2/10) * (1/9) * (8/8).
Simplifying this expression will give you the ultimate probability. Good luck with your calculations!
To find the probability that exactly two of the balls are the same color, we can consider two cases:
Case 1: Two blue balls and one red ball
Case 2: Two red balls and one blue ball
Let's calculate the probability for each case separately and then add them together.
Case 1: Two blue balls and one red ball
First, let's calculate the probability of selecting two blue balls out of the eight blue balls in the bag.
Probability of selecting the first blue ball = 8/10
Probability of selecting the second blue ball = 7/9
Now, let's calculate the probability of selecting one red ball out of the two red balls remaining in the bag.
Probability of selecting the red ball = 2/8
Since the selections are made simultaneously, we multiply the probabilities together.
P(case 1) = (8/10) * (7/9) * (2/8)
Case 2: Two red balls and one blue ball
First, let's calculate the probability of selecting two red balls out of the two red balls in the bag.
Probability of selecting the first red ball = 2/10
Probability of selecting the second red ball = 1/9
Now, let's calculate the probability of selecting one blue ball out of the eight blue balls remaining in the bag.
Probability of selecting the blue ball = 8/8
Again, since the selections are made simultaneously, we multiply the probabilities together.
P(case 2) = (2/10) * (1/9) * (8/8)
To find the total probability, we add the probabilities for both cases.
P(exactly two of the balls are the same color) = P(case 1) + P(case 2)
I will calculate the probabilities for you.
To find the probability of exactly two balls being the same color when three balls are chosen at random, we will consider all the possible outcomes and then determine the favorable outcomes.
First, let's calculate the total number of possible outcomes when selecting three balls from the bag. Since there are 10 balls in total, the number of ways to choose 3 balls can be calculated using combinations:
Total number of outcomes = C(10, 3) = 10! / (3! * (10-3)!) = 120.
Now, let's determine the number of favorable outcomes where exactly two balls are the same color. There are two cases to consider in this scenario:
Case 1: Selecting two blue balls and one red ball.
In this case, we need to choose 2 blue balls from the 8 blue balls available and 1 red ball from the 2 red balls available. The number of favorable outcomes for this case can be calculated using combinations:
Number of favorable outcomes for Case 1 = C(8, 2) * C(2, 1) = (8! / (2! * (8-2)!)) * (2! / (1! * (2-1)!)) = 28 * 2 = 56.
Case 2: Selecting two red balls and one blue ball.
In this case, we need to choose 2 red balls from the 2 red balls available and 1 blue ball from the 8 blue balls available. The number of favorable outcomes for this case can also be calculated using combinations:
Number of favorable outcomes for Case 2 = C(2, 2) * C(8, 1) = (2! / (2! * (2-2)!)) * (8! / (1! * (8-1)!)) = 1 * 8 = 8.
Now, let's calculate the total number of favorable outcomes by summing up the favorable outcomes for each case:
Total number of favorable outcomes = Number of favorable outcomes for Case 1 + Number of favorable outcomes for Case 2
= 56 + 8
= 64.
Finally, we can calculate the probability by dividing the total number of favorable outcomes by the total number of possible outcomes:
Probability = Total number of favorable outcomes / Total number of possible outcomes
= 64 / 120
= 8 / 15.
Therefore, the probability that exactly two of the balls are the same color is 8/15 or approximately 0.53.
P(2blue) = 8/10 * 7/9
P(2red) = 2/10 * 1/9
Either-or probabilities are found by adding the individual probabilities.