1: A box contains 95 pink rubber bands and 90 brown rubber bands. You select a rubber band at random from the box. Find each probability. Write the probability as a fraction in simplest form.

A). Find the theoretical probability of selecting a pink rubber band.

B). Find the theoretical probability of selecting a brown rubber band.

C). You repeatedly choose a rubber band from the box, record the color, and put the rubber band back in the box. The results are shown in the table below. Find the experimental probability of each color based on the table

Pink: 36
Brown: 33

2: The diagram below shows the contents of a jar from which you select marbles at random.

Four red marbles
Seven blue marbles
Five green marbles

A). What is the probability of selecting a red marble, replacing it, and then selecting a blue marble? Show your work.

B). What is the probability of selecting a red marble, setting it aside, and then selecting a blue marble? Show your work.

C). Are the answers to parts A) and B) the same? Why or why not?

The answer to C on 2 is because by replacing the red marble, you've added another number to probability. By setting it aside you're making it simpler. Am I right?

Thanks!~

See Related Questions: Thu, 5-7-15, 5:14 AM.

A) The theoretical probability of selecting a pink rubber band can be found by dividing the number of pink rubber bands (95) by the total number of rubber bands (95+90=185).

P(pink) = 95/185

B) The theoretical probability of selecting a brown rubber band can be found by dividing the number of brown rubber bands (90) by the total number of rubber bands (95+90=185).
P(brown) = 90/185

C) The experimental probability of each color can be found by dividing the number of times each color appears in the table by the total number of trials (36+33=69).
Experimental probability of selecting pink = 36/69
Experimental probability of selecting brown = 33/69

For question 2:

A) The probability of selecting a red marble, replacing it, and then selecting a blue marble can be found by multiplying the probability of selecting a red marble (4/16) by the probability of selecting a blue marble (7/16).
P(red and then blue) = (4/16) * (7/16) = 28/256 Simplify the fraction if needed.

B) The probability of selecting a red marble, setting it aside, and then selecting a blue marble can be found by multiplying the probability of selecting a red marble (4/16) by the probability of selecting a blue marble from the remaining marbles (6/15, as one red marble is already set aside).
P(red and then blue) = (4/16) * (6/15) = 24/240 Simplify the fraction if needed.

C) No, the answers to parts A) and B) are not the same because in part A) the red marble is replaced back into the jar before the second selection, whereas in part B) the red marble is set aside and not available for the second selection.