A bag contains 10 marbles: 1 red, 2 blue, 3 green, and 4 yellow. I reach into the bag and pick
5 marbles at random. Determine the probability that my pick has:
(a) At least 2 yellow marbles.
(b) At least 2 yellow marbles and at least 1 green marble.
(c) All four colors represented.
(d) At least two colors represented.
(e) At most three colors represented.
anyone?
To determine the probabilities, we need to consider the total number of possible outcomes and the number of favorable outcomes for each case. Let's break down each question and calculate the probabilities step by step.
(a) At least 2 yellow marbles:
To calculate the probability of picking at least 2 yellow marbles, we need to consider two scenarios: picking exactly 2 yellow marbles or picking more than 2 yellow marbles.
The total number of possible outcomes is the number of ways to pick 5 marbles out of 10, which is calculated using the combination formula: C(10, 5) = 10! / (5! * (10-5)!)= 252.
For the favorable outcomes, we need to consider two cases:
Case 1: Picking exactly 2 yellow marbles: There are 4 yellow marbles to choose from, and we need to choose 2 out of the 4. So, the number of ways to choose 2 yellow marbles is C(4, 2) = 4! / (2! * (4-2)!) = 6.
Case 2: Picking more than 2 yellow marbles: We can pick 3 or 4 yellow marbles. The number of ways to choose 3 yellow marbles is C(4, 3) = 4! / (3! * (4-3)!) = 4, and the number of ways to choose all 4 yellow marbles is C(4, 4) = 4! / (4! * (4-4)!) = 1.
The total number of favorable outcomes is the sum of the favorable outcomes from both cases: 6 + 4 + 1 = 11.
Therefore, the probability of picking at least 2 yellow marbles is 11/252 ≈ 0.0437.
(b) At least 2 yellow marbles and at least 1 green marble:
To calculate the probability of picking at least 2 yellow marbles and at least 1 green marble, we need to consider the favorable outcomes where both conditions are satisfied.
For the favorable outcomes, we need to consider three cases:
Case 1: Picking exactly 2 yellow marbles and 1 green marble: The number of ways to choose 2 yellow marbles out of 4 and 1 green marble out of 3 is C(4, 2) * C(3, 1) = 6 * 3 = 18.
Case 2: Picking exactly 3 yellow marbles and 1 green marble: The number of ways to choose 3 yellow marbles out of 4 and 1 green marble out of 3 is C(4, 3) * C(3, 1) = 4 * 3 = 12.
Case 3: Picking all 4 yellow marbles and 1 green marble: The number of ways to choose all 4 yellow marbles out of 4 and 1 green marble out of 3 is C(4, 4) * C(3, 1) = 1 * 3 = 3.
The total number of favorable outcomes is the sum of the favorable outcomes from all three cases: 18 + 12 + 3 = 33.
Therefore, the probability of picking at least 2 yellow marbles and at least 1 green marble is 33/252 ≈ 0.1309.