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2 years ago

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A bag contains exactly three red marbles, five yellow marbles and two blue marbles. If three marbles are drawn from the bag with

out replacement, what is the probability that all three will be the same color? Express your answer as a common fraction

1 answer:

Studentka2010 [4]2 years ago

3 0

Answer:

11/120

Step-by-step explanation:

Red = 3
Yellow = 5
Blue = 2

Possible options would be red or yellow as only two blue marbles.

<u>Probability of 3 red:</u>

3/10*2/9*1/8 = 1/120

<u>Probability of 3 yellow:</u>

5/10*4/9*3/8 = 1/12

<u>Probability of same three is:</u>

1/120 + 1/12 = 11/120

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erma4kov [3.2K]

Answer:

$\frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{120}$

Step-by-step explanation:

Given

5 tuples implies that:

$n = 5$

$(h,i,j,k,m)$ implies that:

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Required

How many 5-tuples of integers $(h, i, j, k,m)$ are there such that $n\ge h\ge i\ge j\ge k\ge m\ge 1$

From the question, the order of the integers h, i, j, k and m does not matter. This implies that, we make use of combination to solve this problem.

Also considering that repetition is allowed: This implies that, a number can be repeated in more than 1 location

So, there are n + 4 items to make selection from

The selection becomes:

$^{n}C_r => ^{n + 4}C_5$

$^{n + 4}C_5 = \frac{(n+4)!}{(n+4-5)!5!}$

$^{n + 4}C_5 = \frac{(n+4)!}{(n-1)!5!}$

Expand the numerator

$^{n + 4}C_5 = \frac{(n+4)!(n+3)*(n+2)*(n+1)*n*(n-1)!}{(n-1)!5!}$

$^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{5!}$

$^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{5*4*3*2*1}$

$^{n + 4}C_5 = \frac{(n+4)*(n+3)*(n+2)*(n+1)*n}{120}$

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