There is a bag filled with 5 blue, 6 red and 2 green marbles. A marble is taken at random from the bag, the colour is noted and then it is replaced. Another marble is taken at random. What is the probability of getting exactly 1 blue?
Find the total number of marbles in the bag:
Total marbles = 5 blue + 6 red + 2 green
Total marbles = 13
The problem asks for exactly one blue in 2 draws with replacement. Which means you could draw as follows:
Blue, Not Blue
Not Blue, Blue
The probability of drawing a blue is 5/13, since we replace the marbles in the bag each time.
The probability of not drawing a blue is (6 + 2)/13 = 8/13
And since each of the 2 draws are independent of each other, we multiply the probability of each draw:
Blue, Not Blue = 5/13 * 8/13 =40/169
Not Blue, Blue = 8/13 * 5/13 = 40/169
We add both probabilities since they both count under our scenario:
40/169 + 40/169 = 80/169
Checking our fraction simplification calculator, we see you cannot simplify this fraction anymore.
So our probability stated in terms of a fraction is 80/169
Stated in terms of a decimal, it's 0.4734
Find the total number of marbles in the bag:
Total marbles = 5 blue + 6 red + 2 green
Total marbles = 13
The problem asks for exactly one blue in 2 draws with replacement. Which means you could draw as follows:
Blue, Not Blue
Not Blue, Blue
The probability of drawing a blue is 5/13, since we replace the marbles in the bag each time.
The probability of not drawing a blue is (6 + 2)/13 = 8/13
And since each of the 2 draws are independent of each other, we multiply the probability of each draw:
Blue, Not Blue = 5/13 * 8/13 =40/169
Not Blue, Blue = 8/13 * 5/13 = 40/169
We add both probabilities since they both count under our scenario:
40/169 + 40/169 = 80/169
Checking our fraction simplification calculator, we see you cannot simplify this fraction anymore.
So our probability stated in terms of a fraction is 80/169
Stated in terms of a decimal, it's 0.4734