SOLUTION: An urn contains 7 red marbles, 6 white marbles, and 8 blue marbles marbles. A child randomly selects three (without replacement) from the urn. Round to four decimal places. Find

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Question 1207265: An urn contains 7 red marbles, 6 white marbles, and 8 blue marbles marbles. A child randomly selects three (without replacement) from the urn. Round to four decimal places.
Find the probability all three marbles are the same color.

Find the probability that none of the three marbles are white.
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website!
.
An urn contains 7 red marbles, 6 white marbles, and 8 blue marbles.
A child randomly selects three (without replacement) from the urn.
Round to four decimal places.
(a)Find the probability all three marbles are the same color.
(b) Find the probability that none of the three marbles are white.
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(a) In (a), the probability is the sum of the probabilities of three disjoint events: P = P(all three marbles are red) + P(all three marbles are white) + P(all three marbles are blue) = = + + = 0.0835 (rounded). Here 21 = 7 + 6 + 8 is the total marbles. The structure of the formula is clear, and the structure reflects its meaning. (b) P = = 0.3421 (rounded). Here 15 = 7+8 is the number of all non-white marbles. Again, the structure of the formula is clear, and the structure reflects its meaning.

Solved.

Answer by math_tutor2020(3817) (Show Source):
You can put this solution on YOUR website!

Part 1

Let's find the probability of getting 3 red marbles in a row.
7 red out of 7+6+8 = 21 total
7/21 = probability of getting a red marble
6/20 = probability of a second red marble (no replacement)
5/19 = probability of a third red marble (no replacement)
Note the countdown of numerator and denominator. These countdowns are because of the no replacement.
(7/21)*(6/20)*(5/19) = 1/38 is the probability of selecting 3 red marbles in a row (no replacement)
Another approach is to use the nCr combination formula to find there are 7C3 = 35 ways to pick the 3 red marbles out of 21C3 = 1330 ways to select 3 marbles. Then 35/1330 = 1/38

Now calculate the probability of getting 3 white marbles in a row.
6 white out of 21 total
6/21, 5/20, 4/19 are the probabilities of 1st,2nd,3rd marbles as white if there's no replacement.
(6/21)*(5/20)*(4/19) = 2/133 is the probability of selecting 3 white marbles in a row with no replacement.

Lastly we'll find the probability of getting 3 blue marbles in a row.
8 blue out of 21 total
8/21, 7/20, 6/19 are the probabilities of 1st,2nd,3rd marbles as blue where there's no replacement.
(8/21)*(7/20)*(6/19) = 4/95 is the probability of selecting 3 blue marbles in a row with no replacement.

To summarize so far
P(3 red) = 1/38
P(3 white) = 2/133
P(3 blue) = 4/95
Add up those fractions to get 111/1330 = 0.0835 as the approximate probability of getting 3 of the same color in a row.
Note that those 3 probability values we added represent mutually exclusive events.

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Part 2

7 red + 8 blue = 15 red or blue out of 7+6+8 = 21 marbles total
15/21, 14/20, 13/19 are the probability values for 1st,2nd,3rd marbles as non-white if no replacements are made.

Multiply out those fractions to get 13/38 = 0.3421 approximately.

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Answers:
Probability all three marbles are the same color.
0.0835

Probability that none of the three marbles are white.
0.3421

Both results are approximate.