Question

Assume a jar has five red marbles and two black marbles. Draw out two marbles with and without replacement. Find the requested probabilities. (Enter the probabilities as fractions.)
(a) P(two red marbles)
with replacement
without replacement

Answer 1

With replacement: The total number of marbles is seven. There are 5 red marbles before each of the two draws.
P(red on each draw) = 5/7
P(red on both draws) = (5.7) * (5/7) = you can calculate.

Answer 2

25/49, what about without a replacement?

Answer 3

Without replacement:
P(red on first draw) = 5/7
If a red has been drawn on the first draw, there are now 6 marbles in total and 4 are red. Therefore the probability of red on the second draw is 4/6.
P(red on both draws) = (5/7) * (4/6) = you can calculate.

Answer 4

Thank you so much! Do you mind helping me with two more?
P(one red and one black marble)
with and without replacements

P(red on the first draw and black on the second draw)
with and without replacements

Answer 5

You're welcome :)
P(one red and one black marble) with and without replacements.
With replacements:
P(red first, black second) = (5/7) * (2/7)
P(black first, red second) = (2/7) * (5/7)
The two outcomes are mutually exclusive, therefore the required probability is given by:
(5/7) * (2/7) + (2/7) * (5/7) = you can calculate.

Answer 6

I got 10/49 for P(red first, black second) and it said it was wrong :/
and I got 20/98 for without replacements for both and they were wrong

Answer 7

Am I doing the problems wrong?

Answer 8

"I got 10/49 for P(red first, black second) and it said it was wrong".
Yes it is wrong! Why didn't you use the method that I showed you? You must give the probability based on 'red first, black second' and the alternative 'black first, red second'. The question does not ask for one order of drawing the two colors. Both orders must be considered.

Answer 9

Okay well I'm kind of confused on how to calculate that then

Answer 10

(5/7) * (2/7) + (2/7) * (5/7) = (10/49) + (10/49) = you can calculate.

Answer 11

20/98? Is this for the replacement of P(red first, black second)?

Answer 12

(10/49) + (10/49) = 20/49
That is the probability for:
"P(one red and one black marble) with replacements".
Note that the question allows for 'one black and one red marble' as well as 'one red and one black'. So there are two outcomes that give the required result of one marble of each color.

Answer 13

How do I calculate without replacements?

Answer 14

"P(one red and one black marble) without replacements".
\[\large P(1\ red,\ 1\ black)=\frac{5C1 \times 2C1}{7C2}=\frac{5\times2\times2}{7\times6}=you\ can\ calculate\]

Answer 15

I got 20/42

Answer 16

which simplifies to 10/21.

Answer 17

Okay cool, how about for P(red on the first draw and black on the second draw)

Answer 18

P(red on the first draw and black on the second draw) with replacements.
P(red on first draw) = 5/7
P(black on second draw) = 2/7
P(red on the first draw and black on the second draw) = (5/7) * (2/7)

Answer 19

without replacements?

Answer 20

P(red on the first draw and black on the second draw) without replacements.
P(red on first draw) = 5/7
If red has been drawn on the first draw, there are 6 marbles remaining and 2 of these are black. So P(black on second draw) = 2/6
P(red on the first draw and black on the second draw) = (5/7) * (2/6)

Answer 21

THANK YOU SO MUCH!!!!

Answer 22

You're welcome :)