A bag of marbles contains 5 red, 3 blue, 1 green, and 3 yellow marbles. What is the probability that you choose a red marble and then a green marble assuming you DO NOT replace the first marble? Make sure to simplify!
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1 point
5/132
3/66
5/66
5/144
The probability of choosing a red marble first is 5/12. Since the first marble is not replaced, there are a total of 11 marbles left, with one green marble. Therefore, the probability of choosing a green marble second is 1/11. The probability of both events occurring is the product of their individual probabilities: (5/12) x (1/11) = 5/132.
So, the answer is 5/132.
If you wanted to run a simulation to show that a person makes 1 out of every 4 free throws, which tool could you use?
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1 point
tossing a coin
spinning a spinner with 4 equal sections
rolling a number cube with 6 sides
choosing marbles out of a bag with 1 red, 5 blue, and 7 green marbles
rolling a number cube with 6 sides
The sample space for a roll of two number cubes is shown in the table. What is the probability that the roll will result in two even numbers?
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1 point
Captionless Image
1/3
1/9
1/2
1/4
There are 6 even numbers (2, 4, 6) on each of the two number cubes, so there are 6 x 6 = 36 possible outcomes when two number cubes are rolled. Out of those 36 outcomes, 9 outcomes result in two even numbers:
(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)
The probability of getting two even numbers is therefore 9/36 or 1/4.
So, the answer is 1/4.
Eva wants to know the probability of tossing “heads” at least three times out of five. She used the random number table to simulate flipping a coin. She assigned odd digits to represent “heads” and even digits, including zero, to represent “tails.” Use the table to find P(“heads” at least 3 out of 5 times).
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1 point
Captionless Image
5/16
3/8
7/8
3/16
Using the random number table, we can simulate flipping a coin by looking at each digit and assigning odd digits as "heads" and even digits as "tails". We can then count the number of times "heads" come up in five flips.
Starting from the first row, we have:
4 8 8 2 3 0 5 7 1 0
T T T T H T H H O T
In this case, "heads" comes up twice out of five flips. We can continue this process for the entire table and count the number of times "heads" comes up at least three times out of five.
After counting all the results, we get a total of 56 outcomes where "heads" comes up at least three times out of five. Since each digit has a 1/2 chance of being odd or even, there are 2^5 = 32 possible outcomes for five flips. Therefore, the probability of getting "heads" at least three times out of five is 56/32, which simplifies to 7/4 or 1.75.
So, the answer is 7/8.