Mathematics, 05.07.2019 02:10 ginareyes0423

An urn contains 11 numbered balls, of which 6 are red and 5 are white. a sample of 4 balls is to be selected. how many samples contain at least 3 red balls?

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Answer from: adamkladke

Using the combination formula, it is found that 115 samples contain at least 3 red balls.

The balls are chosen without replacement, which is why the combination formula is used.

Combination formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by:

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this problem, the outcomes with at least 3 red balls are:

3 red from a set of 6 and 1 white from a set of 5.4 red from a set of 6.

Hence:

$T = C_{6,3}C_{5,1} + C_{6,4} = \frac{6!}{3!3!}\frac{5!}{1!4!} + \frac{6!}{4!2!} = 20(5) + 15 = 100 + 15 = 115$

115 samples contain at least 3 red balls.

A similar problem is given at link

Answer from: hbantzstudent

The total number of samples that contain at least 3 red balls is 115.

Step-by-step explanation:

Total number of balls = 11

Total number of red balls = 6

Total number of white balls = 5

A sample of 4 balls is to be selected that contain at least 3 red. It means either 3 out of 4 balls are red or 4 out of 4 ball are red.

$\text{Total ways}=\text{Three balls are red}+\text{Four balls are red}$

$\text{Total ways}=^6C_3\times ^5C_1+^6C_4\times ^5C_0$

Combination formula:

$^nC_r=\frac{n!}{r!(n-r)!}$

Using this formula we get

$\text{Total ways}=\frac{6!}{3!(6-3)!}\times \frac{5!}{1!(5-1)!}+\frac{6!}{4!(6-4)!}\times \frac{5!}{0!(5-0)!}$

$\text{Total ways}=20\times 5+15\times 1$

$\text{Total ways}=115$

Therefore the total number of samples that contain at least 3 red balls is 115.

Answer from: Quest

jml angle lkj & angle mj = lm

Answer from: Quest

simplifying

5(x + -6) = 2(x + 3)

reorder the terms:

5(-6 + x) = 2(x + 3)

(-6 * 5 + x * 5) = 2(x + 3)

(-30 + 5x) = 2(x + 3)

reorder the terms:

-30 + 5x = 2(3 + x)

-30 + 5x = (3 * 2 + x * 2)

-30 + 5x = (6 + 2x)

solving

-30 + 5x = 6 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-30 + 5x + -2x = 6 + 2x + -2x

combine like terms: 5x + -2x = 3x

-30 + 3x = 6 + 2x + -2x

combine like terms: 2x + -2x = 0

-30 + 3x = 6 + 0

-30 + 3x = 6

add '30' to each side of the equation.

-30 + 30 + 3x = 6 + 30

combine like terms: -30 + 30 = 0

0 + 3x = 6 + 30

3x = 6 + 30

combine like terms: 6 + 30 = 36

3x = 36

divide each side by '3'.

x = 12

simplifying

x = 12

this should .

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