a jar of marbles contains 3 red marbles 5 blue marbles 4 green marbles someone take 3 marbles from the jar one at a time without replacement what is the probability that someone picks a red marble then a blue marble then a green marble?
Looks like the order is important, so
prob(as stated) = (3/12)(5/11)(4/10) = ....
3/12 * 5/11 * 4/10 = ?
To find the probability of someone picking a red marble, then a blue marble, and finally a green marble without replacement, we need to calculate the probability of each event occurring one after the other.
Step 1: Probability of picking a red marble
Initially, there are 3 red marbles out of a total of 12 marbles in the jar. Therefore, the probability of picking a red marble on the first pick is 3/12.
Step 2: Probability of picking a blue marble
After the first marble is picked without replacement, there are now 11 marbles remaining in the jar, of which 5 are blue. So the probability of picking a blue marble on the second pick is 5/11.
Step 3: Probability of picking a green marble
After the second marble is picked without replacement, there are 10 marbles left, out of which 4 are green. Hence, the probability of picking a green marble on the final pick is 4/10.
To find the overall probability of these events occurring together, we multiply the individual probabilities:
P(Red, Blue, Green) = P(Red) * P(Blue) * P(Green)
= (3/12) * (5/11) * (4/10)
= 60 / 1320
= 1 / 22
Therefore, the probability of someone picking a red marble, then a blue marble, and finally a green marble is 1/22, or approximately 0.045.