A old man lives in a apartment containing 2 rooms. Each day before going to work he enters any one room randomly, picks up a bag and leaves home. One of the rooms contains 3 blue, 4 green and 5 red bags and the other contains 2 blue, 1 green and 3 red bags. What is the probability that he takes a green bag to work?Options1/71/41/31/5
Question
A old man lives in a apartment containing 2 rooms. Each day before going to work he enters any one room randomly, picks up a bag and leaves home. One of the rooms contains 3 blue, 4 green and 5 red bags and the other contains 2 blue, 1 green and 3 red bags. What is the probability that he takes a green bag to work?Options1/71/41/31/5
Solution
To solve this problem, we first need to calculate the total number of bags in each room and the total number of green bags in each room.
In the first room, there are 3 blue + 4 green + 5 red = 12 bags in total, with 4 of them being green.
In the second room, there are 2 blue + 1 green + 3 red = 6 bags in total, with 1 of them being green.
Since the man chooses a room randomly, the probability of him entering either room is 1/2.
So, the probability of him picking a green bag from the first room is (1/2) * (4/12) = 1/6.
And the probability of him picking a green bag from the second room is (1/2) * (1/6) = 1/12.
Adding these two probabilities together gives us the total probability of the man picking a green bag, which is 1/6 + 1/12 = 1/4.
So, the correct answer is 1/4.
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