There is a bag filled with 4 blue, 3 red and 5 green marbles.
A marble is taken at random from the bag, the colour is noted and then it is not replaced.
Another marble is taken at random.
What is the probability of getting exactly 1 blue?
To solve this probability problem, we need to consider the total number of marbles and the number of blue marbles.
1. First, determine the total number of marbles in the bag. In this case, the bag contains 4 blue marbles, 3 red marbles, and 5 green marbles, so there are a total of 4 + 3 + 5 = 12 marbles.
2. Next, let's consider the first marble that is picked. Since we want to calculate the probability of getting exactly 1 blue marble, there are two possibilities for the first marble: either it is blue or it is not.
- Case 1: The first marble is blue. There are 4 blue marbles in the bag, so the probability of selecting a blue marble on the first pick is 4/12.
- Case 2: The first marble is not blue. There are 8 non-blue marbles in the bag (3 red and 5 green), so the probability of not selecting a blue marble on the first pick is 8/12.
3. After the first marble is picked and not replaced, there are now 11 marbles left in the bag. We want to calculate the probability of selecting a blue marble on the second pick.
- Case 1: If the first marble was blue, there are now only 3 blue marbles left out of the 11 in the bag. So the probability of selecting a blue marble on the second pick, given that the first marble was blue, is 3/11.
- Case 2: If the first marble was not blue, there are still 4 blue marbles left out of the 11 in the bag. So the probability of selecting a blue marble on the second pick, given that the first marble was not blue, is 4/11.
4. Finally, we need to account for the fact that the first marble could be blue or not blue. Thus, we need to calculate the overall probability by multiplying the probabilities from each case:
Overall probability = (Probability of Case 1 * Probability of Case 1's outcome) + (Probability of Case 2 * Probability of Case 2's outcome)
Overall probability = (4/12 * 3/11) + (8/12 * 4/11)
Performing the calculations, the overall probability of getting exactly 1 blue marble is approximately 0.3636, or 36.36%.
To find the probability of getting exactly 1 blue marble, we need to consider the total number of possibilities and the number of favorable outcomes.
Total number of possibilities:
When the first marble is taken out, there are 12 marbles in the bag (4 blue + 3 red + 5 green). After removing one marble, we are left with 11 marbles. Therefore, there are 11 possible outcomes for the second marble.
Number of favorable outcomes:
To find exactly 1 blue marble, we have two cases to consider:
Case 1: The first marble is blue and the second marble is not blue.
In this case, there are 4 blue marbles to choose from initially, and after removing one blue marble, we are left with 11 marbles. Therefore, there are 4 * 11 = 44 favorable outcomes.
Case 2: The first marble is not blue and the second marble is blue.
In this case, there are 8 non-blue marbles initially (3 red + 5 green). Then, after removing one non-blue marble, there are 8 possible outcomes for the first marble. For the second marble, we have 4 blue marbles left. Therefore, there are 8 * 4 = 32 favorable outcomes.
Total number of favorable outcomes: 44 + 32 = 76
Now, we can calculate the probability by dividing the number of favorable outcomes by the total number of possibilities:
Probability of getting exactly 1 blue marble = Number of favorable outcomes / Total number of possibilities
Probability = 76 / 11
Probability ≈ 0.7273
Therefore, the probability of getting exactly 1 blue marble is approximately 0.7273, or 72.73%.
possible outcomes:
B,B ---- (4/12)(3/11) = 12/132
B, not blue = (4/12)(8/11) = 32/132
not blue, B = (8/12)(4/11) = 32/132
notblue, notblue = (8/12)(7/11) = 56/132
Note 12/132 + 32/132 + 32/132 + 56/132 = 132/132 = 1, as it should
so prob(exactly 1 blue out of 2 picks) = 32/132 + 32/132 = 16/33