A bag contains 10 red chips and 12 blue chips. Two chips are selected randomly without replacement from the bag.
What is the probability that the two chips are the same color? Round your answer to 3 decimal places..
Help me find the answer.
Is there another way to answer it without the tree method?
You could pick either 2 reds or 2 blues
prob(2 reds) = (10/22)(9/21) = 15/77
prob(2 blues) = (12/22)(11/21) = 2/7
prob(your event) = 15/77 + 2/7 = 37/77
Yes, there is another way to solve this problem without using a tree diagram. We can calculate the probability using combinations.
To find the probability that both chips are the same color, we need to consider two cases: the probability of selecting two red chips and the probability of selecting two blue chips.
Case 1: Selecting two red chips:
In this case, we need to choose 2 chips from the 10 red chips available. The total number of ways to choose 2 chips from 10 is given by the combination formula: C(10, 2) = (10!)/(2!(10-2)!) = 45.
Case 2: Selecting two blue chips:
Similar to Case 1, we need to choose 2 chips from the 12 blue chips available. The total number of ways to choose 2 chips from 12 is given by the combination formula: C(12, 2) = (12!)/(2!(12-2)!) = 66.
The total number of possible combinations to select 2 chips from 22 (10 red + 12 blue) chips is given by C(22, 2) = (22!)/(2!(22-2)!) = 231.
Therefore, the probability of selecting two chips of the same color is (45 + 66)/231 ≈ 0.558 (rounded to 3 decimal places).
So, the probability that the two chips selected are the same color is approximately 0.558.
Yes, there is another way to find the answer without using the tree method. We can approach this problem using combinations.
To determine the probability that the two chips are the same color, we need to find two scenarios: when both chips are red and when both chips are blue. Then we will add these probabilities together.
To calculate the probability that both chips are red, we need to calculate the number of ways to choose 2 red chips from the 10 red chips in the bag, divided by the total number of ways to choose 2 chips from all the chips in the bag. This can be expressed as:
P(Red, Red) = (number of ways to choose 2 red chips) / (number of ways to choose 2 chips)
The number of ways to choose 2 red chips can be calculated using combinations. We can use the formula C(n, r) = n! / (r!(n-r)!), where n is the total number of items and r is the number of items you want to choose. In this case, n = 10 and r = 2.
So, the number of ways to choose 2 red chips is C(10, 2) = 10! / (2!(10-2)!) = 45.
The total number of ways to choose 2 chips from all the chips in the bag can also be calculated using combinations. In this case, we have a total of 22 chips (10 red + 12 blue), so n = 22 and r = 2.
So, the number of ways to choose 2 chips from the bag is C(22, 2) = 22! / (2!(22-2)!) = 231.
Now, we can calculate the probability:
P(Red, Red) = 45 / 231 = 0.195
Similarly, we can calculate the probability that both chips are blue:
P(Blue, Blue) = (number of ways to choose 2 blue chips) / (number of ways to choose 2 chips)
Using combinations, the number of ways to choose 2 blue chips is C(12, 2) = 12! / (2!(12-2)!) = 66.
So, P(Blue, Blue) = 66 / 231 = 0.286
Finally, we add these probabilities together to get the probability that the two chips are the same color:
P(Same color) = P(Red, Red) + P(Blue, Blue) = 0.195 + 0.286 ≈ 0.481
Therefore, the probability that the two chips are the same color is approximately 0.481, rounded to 3 decimal places.