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lana [24]
3 years ago
7

Probability Question

Mathematics
2 answers:
trapecia [35]3 years ago
8 0

Answer:

39/812

Step-by-step explanation:

The total no. of balls is 6+7+8+9 = 30.

There are 4 possible outcomes that fits "all 3 balls taken are of the same color": 1. all red balls/2.all green balls / 3. all blue balls/4. all white balls. Either one will do. So the required probability is the sum of probability that these 4 outcomes happen.

First lets find the probability of taking out all red balls.

There are total 30 balls, but only 6 are red, so in the first try, the probability of taking out a red ball is 6/30.

In the 2nd time, since you have already taken out one red ball, the no. of red balls left is 5, and the total no. of balls is 29. So the probability of taking out another red ball is 5/29.

This repeats for the 3rd time. The probability of taking out the 3rd red ball is 4/28.

These 3 times all have to happen in order, so the total probability is

6/30 x 5/29 x 4/28

=1/203

Now find the probability of taking out all green balls. The process is same from finding the probability of taking out all red balls, just that replace 6 to 7 instead.

So, the probability of taking out all green balls is

7/30 x 6/29 x 5/28

=1/116

Repeat again for blue balls and white balls.

Probability of taking out all blue balls = 8/30 x 7/29 x 6/28 = 2/145

Probability of taking out all white balls= 9/30 x 8/29 x7/28 =3/145

Now add all 4 fractions up.

1/203 + 1/116 + 2/145 + 3/145

=39/812

By the way, another method is to use combination "nCr". You will also get the same answer.

n: number of items

r: number of items being chosen at a time

(Probability of all red balls + Probability of all green balls + probability of all blue balls + probability of all white balls)÷Total no. of combinations - which is random 3 balls out of 30.

(6C3 + 7C3 + 8C3 +9C3) / (30C3)

= 39/812

strojnjashka [21]3 years ago
5 0

Answer:  

39/812.

Step-by-step explanation:

There are a total of 6+7+8+9 = 30 balls.

Prob(3 red balls are taken) =  6/30 * 5/29 * 4/28 = 1/203

Prob(3 green balls are taken) =  7/30 * 6/29 * 5/28 = 1/116.

Prob(3 blue balls are taken) =  8/30 * 7/29 * 6/28 = 2/145.

Prob(3 white balls are taken) = 9/30 * 8/29 * 7/28 = 4/145.

So the probability of 3 balls od the same colour is the sum of these above probabilities =  39/812.

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A product can be made in sizes huge, average and tiny which yield a net unit profit of $14, $10, and$5, respectively. Three cent
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Answer:

The model is:

z = 14* X₁₁ + 14*X₁₂ + 5*X₁₃ + 10*X₂₁ + 10*X₂₂ + 10*X₂₃ + 5*X₃₁ + 5*X₃₂ + 5*X₃₃    to maximize

Subject to:

First center               X₁₁  +  X₂₁  + X₃₁  ≤  550

Second center         X₁₂  +  X₂₂  + X₃₂  ≤ 750

Third center               X₁₃  + X₂₃  + X₃₃  ≤ 275                  

22* X₁₁  + 16* X₂₁  + 9*X₃₁     ≤   11000

22* X₁₂  + 16* X₂₂  + 9*X₃₂   ≤   2700

22*X₁₃  + 16* X₂₃  +  9*X₃₃  ≤  3400

X₁₁  +  X₁₂  + X₁₃  ≤  710

X₂₁  + X₂₂ + X₂₃  ≤  900

X₃₁ + X₃₂ + X₃₃  ≤  350

2700*(X₁₁  +  X ₂₁  + X₃₁)  -  11000*(X₁₂ + X₂₂ + X₃₂) = 0

3400*(X₁₁  +  X ₂₁  + X₃₁) - 11000*( ( X₁₃ + X₂₃ + X₃₃) = 0

Xij >= 0

Step-by-step explanation:

Let´s call Xij   product size i produced in center j

According to this, we get the following set of variable

X₁₁    product size huge produced in center 1

X₁₂    product size huge produced in center 2

X₁₃   product size huge produced in center 3

X₂₁   product size average produced in center 1

X₂₂   product size average produced in center 2

X₂₃   product size average produced in center 3

X₃₁  product size-tiny produced in center 1

X₃₂ product size-tiny produced in center 2

X₃₃ product size-tiny produced in center 3

Then Objective function is

z = 14* X₁₁ + 14*X₁₂ + 5*X₁₃ + 10*X₂₁ + 10*X₂₂ + 10*X₂₃ + 5*X₃₁ + 5*X₃₂ + 5*X₃₃

Constrains

Center capacity

1.-   First center               X₁₁  +  X₂₁  + X₃₁  ≤  550

2.-   Second center         X₁₂  +  X₂₂  + X₃₂  ≤ 750

3.- Third center               X₁₃  + X₂₃  + X₃₃  ≤ 275

Water available

1.-  22* X₁₁  + 16* X₂₁  + 9*X₃₁     ≤   11000

2.-  22* X₁₂  + 16* X₂₂  + 9*X₃₂   ≤   2700

3.-   22*X₁₃  + 16* X₂₃  +  9*X₃₃  ≤  3400

Demand constrain

Product huge

X₁₁  +  X₁₂  + X₁₃  ≤  710

Product average

X₂₁  + X₂₂ + X₂₃  ≤  900

Product tiny

X₃₁ + X₃₂ + X₃₃  ≤  350

Fraction SP/CC must be the same

First and second centers  fraction SP/CC  

(X₁₁  +  X ₂₁  + X₃₁)/ 11000   =  (X₁₂ + X₂₂ + X₃₂)/ 2700

2700*(X₁₁  +  X ₂₁  + X₃₁)  -  11000*(X₁₂ + X₂₂ + X₃₂) = 0

First and third centers  fraction SP/CC  

(X₁₁  +  X ₂₁  + X₃₁)/ 11000   = ( X₁₃ + X₂₃ + X₃₃)/ 3400

3400*(X₁₁  +  X ₂₁  + X₃₁) - 11000*( ( X₁₃ + X₂₃ + X₃₃) = 0

The model is:

z = 14* X₁₁ + 14*X₁₂ + 5*X₁₃ + 10*X₂₁ + 10*X₂₂ + 10*X₂₃ + 5*X₃₁ + 5*X₃₂ + 5*X₃₃

Subject to:

First center               X₁₁  +  X₂₁  + X₃₁  ≤  550

Second center         X₁₂  +  X₂₂  + X₃₂  ≤ 750

Third center               X₁₃  + X₂₃  + X₃₃  ≤ 275                  

22* X₁₁  + 16* X₂₁  + 9*X₃₁     ≤   11000

22* X₁₂  + 16* X₂₂  + 9*X₃₂   ≤   2700

22*X₁₃  + 16* X₂₃  +  9*X₃₃  ≤  3400

X₁₁  +  X₁₂  + X₁₃  ≤  710

X₂₁  + X₂₂ + X₂₃  ≤  900

X₃₁ + X₃₂ + X₃₃  ≤  350

2700*(X₁₁  +  X ₂₁  + X₃₁)  -  11000*(X₁₂ + X₂₂ + X₃₂) = 0

3400*(X₁₁  +  X ₂₁  + X₃₁) - 11000*( ( X₁₃ + X₂₃ + X₃₃) = 0

Xij >= 0

6 0
3 years ago
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