An urn contains three yellow,four green and five blue balls. Two balls are randomly drawn without replacement. Find the probability of the following events: a.both balls are blue. b.the first ball is green and the second yellow. c.there is one green and one yellow ball
To find the probabilities of the given events, we need to use the concept of combinations.
The total number of balls in the urn is:
Total = 3 yellow + 4 green + 5 blue = 12 balls
a. To find the probability that both balls are blue, we need to calculate the number of ways we can select 2 blue balls out of the 5 available blue balls.
Number of ways to select 2 blue balls = C(5,2) = 5! / (2!(5-2)!) = 10
Thus, the probability of selecting both blue balls is: P(both blue) = Number of favorable outcomes / Total number of outcomes
= 10 / C(12,2)
The total number of outcomes is the number of ways to select 2 balls out of the 12 available balls, which is given by:
C(12,2) = 12! / (2!(12-2)!) = 66
Therefore, P(both blue) = 10 / 66 = 5 / 33 ≈ 0.152
b. To find the probability that the first ball is green and the second ball is yellow, we need to calculate the number of ways we can select 1 green ball out of 4 available green balls and 1 yellow ball out of 3 available yellow balls.
Number of ways to select 1 green ball out of 4 green balls = C(4,1) = 4
Number of ways to select 1 yellow ball out of 3 yellow balls = C(3,1) = 3
Thus, the number of favorable outcomes is: Number of ways to select 1 green ball * Number of ways to select 1 yellow ball = 4 * 3 = 12
Therefore, P(first ball is green and second ball is yellow) = Number of favorable outcomes / Total number of outcomes
= 12 / C(12,2)
Using the above-mentioned calculations, we can find the probability.
c. To find the probability that there is one green and one yellow ball, we need to calculate the number of ways we can select 1 green ball out of 4 available green balls and 1 yellow ball out of 3 available yellow balls.
Number of ways to select 1 green ball out of 4 green balls = C(4,1) = 4
Number of ways to select 1 yellow ball out of 3 yellow balls = C(3,1) = 3
Thus, the number of favorable outcomes is: Number of ways to select 1 green ball * Number of ways to select 1 yellow ball = 4 * 3 = 12
Therefore, P(one green and one yellow) = Number of favorable outcomes / Total number of outcomes
= 12 / C(12,2)
Using the above-mentioned calculations, we can find the probability.
To find the probabilities of these events, we need to consider the total number of possible outcomes and the number of favorable outcomes for each event.
a. Both balls are blue:
Total number of balls = 3 yellow + 4 green + 5 blue = 12
Total number of ways to choose 2 balls without replacement = 12C2 = (12! / (2! * (12-2)!) = 66
Number of favorable outcomes (choosing 2 blue balls) = 5C2 = (5! / (2! * (5-2)!) = 10
Probability of both balls being blue = Number of favorable outcomes / Total number of outcomes
P(Both balls are blue) = 10/66 = 5/33
b. The first ball is green and the second is yellow:
Total number of balls = 3 yellow + 4 green + 5 blue = 12
Total number of ways to choose 2 balls without replacement = 12C2 = 66
Number of favorable outcomes (choosing 1 green ball and 1 yellow ball) = 4C1 * 3C1 = (4! / (1! * (4-1)!) * 3! / (1! * (3-1)!) = 4 * 3 = 12
Probability of the first ball being green and the second being yellow = Number of favorable outcomes / Total number of outcomes
P(First ball is green and the second is yellow) = 12/66 = 2/11
c. There is one green and one yellow ball:
Total number of balls = 3 yellow + 4 green + 5 blue = 12
Total number of ways to choose 2 balls without replacement = 12C2 = 66
Number of favorable outcomes (choosing 1 green ball and 1 yellow ball) = 4C1 * 3C1 = (4! / (1! * (4-1)!) * 3! / (1! * (3-1)!) = 4 * 3 = 12
Number of favorable outcomes (choosing 1 yellow ball and 1 green ball) = 3C1 * 4C1 = (3! / (1! * (3-1)!) * 4! / (1! * (4-1)!) = 3 * 4 = 12
Total number of favorable outcomes = Number of favorable outcomes (1 green and 1 yellow) + Number of favorable outcomes (1 yellow and 1 green)
= 12 + 12 = 24
Probability of having one green and one yellow ball = Number of favorable outcomes / Total number of outcomes
P(One green and one yellow ball) = 24/66 = 4/11