There is a bag filled with 3 blue and 4 red 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 at least 1 red?
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To find the probability of getting at least 1 red marble, we can calculate the probability of getting no red marbles and subtract it from 1.
First, let's consider the probability of getting no red marbles. When the first marble is drawn, there are a total of 7 marbles in the bag. Since there are 4 red marbles, the probability of not getting a red marble on the first draw is 4/7.
After the first marble is drawn, there are 6 marbles left in the bag, with 3 of them being blue and 3 of them being red. So when the second marble is drawn, the probability of not getting a red marble is 3/6 = 1/2.
To find the probability of not getting a red marble on both draws, we multiply the probabilities of each individual draw: (4/7) * (1/2) = 2/7.
Now, to find the probability of getting at least 1 red marble, we subtract the probability of getting no red marbles from 1: 1 - 2/7 = 5/7.
Therefore, the probability of getting at least 1 red marble is 5/7.
P(some red) = 1 - P(both blue)
= 1 - (3/7 * 2/6)