A bowl contains only red marbles, blue marbles and green marbles. The probability of selecting a red marble from the bowl is 3/13. The probability of selecting a blue marble from the bowl is 2/5. There are fewer than 100 marbles in the bowl. What is the probability of selecting, at random and without replacement, a green marble and then a red marble from the bowl on the first two selections?
Express your answer as a common fraction.
Say there are 65 marbles in the bag. Then there will be 15 red marbles, 26 blue marbles, and 24 green marbles. so without replacement the probability of receiving a green marble and then a red marble would be (24/65)(15/64) which is 9/104
probability of choosing a red marble is 1/5 also white marble is 3/10 what is the blue marble probability?
Well, you know what they say, "The early bird gets the worm, but the second mouse gets the cheese!" Let's calculate the probability of selecting a green marble and then a red marble.
To begin with, the probability of selecting a green marble at random is unknown. However, let's give it a name for now and call it "G." Moving on to the second selection, the probability of choosing a red marble, given that we've already grabbed a green one, is 2/12 since there is one fewer marble and two fewer red marbles.
Now, to calculate the probability of selecting a green marble and then a red marble, we multiply the two probabilities together:
(G probability) x (R probability) = (unknown probability) x (2/12).
Now, we know that the probability of selecting a green marble and then a red marble is 3/13. Therefore, we can set up the equation:
(unknown probability) x (2/12) = 3/13.
To find the unknown probability, we isolate it by dividing both sides of the equation by 2/12:
(unknown probability) = (3/13) / (2/12).
= (3/13) x (12/2).
= (3/13) x 6.
= 9/13.
So, the probability of selecting, at random and without replacement, a green marble and then a red marble from the bowl is 9/13. Hope that brings a smile to your face!
To find the probability of selecting a green marble and then a red marble, we need to multiply the individual probabilities together.
First, let's find the probability of selecting a green marble. We are given that the probability of selecting a red marble is 3/13 and the probability of selecting a blue marble is 2/5. Therefore, the probability of selecting a green marble can be found by subtracting the sum of these two probabilities from 1 (since the total probability of selecting a marble must equal 1).
Probability of selecting a green marble = 1 - (Probability of selecting a red marble + Probability of selecting a blue marble)
Probability of selecting a green marble = 1 - (3/13 + 2/5)
To simplify the expression, we need to find a common denominator for 13 and 5, which is 65.
Probability of selecting a green marble = 1 - (15/65 + 26/65)
Probability of selecting a green marble = 1 - (41/65)
Probability of selecting a green marble = 24/65
Now that we have the probability of selecting a green marble, we need to find the probability of selecting a red marble next. Since we are selecting without replacement, the number of marbles in the bowl reduces by one after the first selection.
Probability of selecting a green marble and then a red marble = probability of selecting a green marble × probability of selecting a red marble (after the green marble is selected)
Probability of selecting a green marble and then a red marble = (24/65) × (3/12)
To simplify this expression, we can cancel out the common factors:
Probability of selecting a green marble and then a red marble = (2/5) × (1/4)
Probability of selecting a green marble and then a red marble = 2/20
Probability of selecting a green marble and then a red marble = 1/10
Therefore, the probability of selecting a green marble and then a red marble from the bowl on the first two selections is 1/10.
3/13 = .23 for red marble
2/5 = .40 for blue marble
?? for green marble.
How many fewer than 100?
Probability of both/all events occurring is found by multiplying the probabilities of the individual events.