a bag contains 4 red chips, 2 blue chips and 5 green chips. A chip is randomly removed from the bag, and then without replacing the first chip, a second chip is removed. What is the probability that both of the chips selected from the bag were blue?
2/11 for first blue chip and 1/10 for the second.
To find probability both/all events would occur, multiply the individual probabilities.
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To find the probability that both chips selected from the bag were blue, we need to find the probability of selecting a blue chip on the first draw and then selecting another blue chip on the second draw without replacing the first chip.
Step 1: Determine the total number of chips in the bag.
There are a total of 4 + 2 + 5 = 11 chips in the bag.
Step 2: Calculate the probability of selecting a blue chip on the first draw.
There are 2 blue chips out of the total 11 chips in the bag, so the probability of selecting a blue chip on the first draw is 2/11.
Step 3: Calculate the probability of selecting a blue chip on the second draw without replacement.
After one blue chip has been removed, there are 1 blue chip remaining out of 10 total chips. Therefore, the probability of selecting a blue chip on the second draw without replacement is 1/10.
Step 4: Calculate the overall probability of selecting two blue chips.
Since we want both events (blue chip on the first draw and blue chip on the second draw) to occur, we multiply the probabilities calculated in Step 2 and Step 3.
Probability = (2/11) * (1/10) = 2/110 = 1/55.
Therefore, the probability that both chips selected from the bag were blue is 1/55.
To find the probability of selecting two blue chips, we need to calculate the ratio of the favorable outcomes to the total number of possible outcomes.
First, let's find the total number of chips in the bag:
Total number of chips = number of red chips + number of blue chips + number of green chips
Total number of chips = 4 + 2 + 5 = 11
We need to consider two scenarios in order to calculate the probability:
1. Selecting the first blue chip and then selecting the second blue chip.
2. Selecting the first non-blue chip (red or green) and then selecting the second blue chip.
Scenario 1:
For the first blue chip, there are 2 blue chips out of a total of 11 chips.
So the probability of selecting the first blue chip is 2/11.
After selecting the first blue chip, there will be 10 chips remaining, with 1 blue chip remaining.
So for the second blue chip, the probability is 1/10.
Therefore, the probability of scenario 1 is (2/11) * (1/10) = 2/110 = 1/55.
Scenario 2:
For the first non-blue chip, there are (4 red chips + 5 green chips) out of a total of 11 chips.
So the probability of selecting the first non-blue chip is (4 + 5)/11 = 9/11.
After selecting the first non-blue chip, there will be 10 chips remaining, with still 2 blue chips remaining.
So for the second blue chip, the probability is 2/10.
Therefore, the probability of scenario 2 is (9/11) * (2/10) = 18/110 = 9/55.
Finally, we add up the probabilities of both scenarios to get the total probability of selecting two blue chips:
Total probability = probability of scenario 1 + probability of scenario 2
Total probability = 1/55 + 9/55
Total probability = 10/55 or simplified as 2/11.
Therefore, the probability that both chips selected from the bag will be blue is 2/11.