Liam has a bag containing five red, two blue, five green, six yellow and two white marbles that are all the same size and shape. what is the probability of randomly choosing a green marble on the first pick, not replacing it, and then randomly choosing a green marble on the second pick?
5 green
20 total
first green = 5/20 = 1/4
second green = 4/19
(1/4)(4/19) = 1/19
You deposit $7,900 in a money-market account that pays an annual interest rate of 4.3%. The interest is compounded quarterly. How much money will you have after 3 years?
To find the probability of randomly choosing a green marble on the first pick and then not replacing it, we first need to determine the total number of marbles in the bag after the first pick.
Initially, there are 5 red, 2 blue, 5 green, 6 yellow, and 2 white marbles, making a total of 20 marbles in the bag.
For the first pick, we want to choose a green marble. The number of green marbles is 5. So, the probability of choosing a green marble on the first pick is 5/20 or 1/4.
After the first pick, we do not replace the marble, so the total number of marbles in the bag decreases by 1.
Now, we need to find the probability of choosing a green marble on the second pick. Since the green marble from the first pick was not replaced, there are now 4 green marbles left in the bag (the first green marble chosen is no longer available).
The reduced total number of marbles is 20 - 1 = 19.
For the second pick, we want to choose a green marble again. The number of green marbles is now 4. So, the probability of choosing a green marble on the second pick is 4/19.
To find the combined probability, we multiply the individual probabilities:
P(1st green and 2nd green) = P(1st green) * P(2nd green)
= (1/4) * (4/19)
= 4/76
= 1/19
Therefore, the probability of randomly choosing a green marble on the first pick and not replacing it, and then randomly choosing a green marble on the second pick is 1/19.