A bag contains 8 green marbles and 4 blue marbles. If two marbles are drawn in succession without replacing them in the bag, what is the probability of drawing a green marble on the second draw?
it will depend on what happened in the first draw.
case1 , first is green, 2nd is green
prob = (8/12)(7/11) = 14/33
case2 , first is not green, 2nd is green
prob = (4/12)(8/11) = 8/33
prob(of your event) = 14/33 + 8/33 = 22/33 = 2/3
To find the probability of drawing a green marble on the second draw, we need to consider the number of favorable outcomes (drawing a green marble) and the number of total outcomes.
First, let's determine the total number of marbles in the bag. There are 8 green marbles and 4 blue marbles, so the total number of marbles in the bag is 8 + 4 = 12.
For the first draw, we have 12 marbles in total, so the probability of drawing a green marble on the first draw is 8/12.
After the first draw, since the marble is not replaced, there are now 11 marbles left in the bag. Out of these 11 marbles, 7 are green (one green marble was already drawn), and 4 are blue.
Therefore, for the second draw, the probability of drawing a green marble is 7/11.
To find the probability of both events happening (drawing a green marble on the first AND second draw), we multiply the probabilities: (8/12) * (7/11).
Simplifying this expression, we get:
(8/12) * (7/11) = 56/132 = 14/33.
So the probability of drawing a green marble on the second draw without replacement is 14/33.