SOLUTION: There are 8 white and 8 black marbles in a basket. We pick out 4 marbles one by one without returning them. Situation A - first and second marbles have at least one white marble. S

Question 1195021: There are 8 white and 8 black marbles in a basket. We pick out 4 marbles one by one without returning them. Situation A - first and second marbles have at least one white marble. Situation B - second and third marble have at least one of them black. Situation C - third and forth marble are both white. Find the probability of event D = not(notA ∪ B) ∩ C
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

The first task is to somehow decipher the notation not(notA ∪ B) ∩ C

Use De Morgan's Law to turn the not(notA ∪ B) portion into not(notA) ∩ not B which further simplifies to A ∩ not B. The two "not" keywords cancel each other out.

In short: not(notA ∪ B) is the same as A ∩ not B

The original set not(notA ∪ B) ∩ C becomes A ∩ not B ∩ C

What does this mean? It means we'll have cases A and C to be true, but case B is false. The symbol ∩ is the intersection symbol to basically represent the keyword "and".

We can think of it like this:
(A is true) and (B is false) and (C is true)

Since case C is true, we know the third and fourth marbles are both white.

We found that case B isn't true. The opposite of "second and third marble have at least one of them black" is "second and third marbles are both white".
It might help to think of a number line.
X = number of black marbles
"At least 1 black" means (ie 1 or more black)
So we could have X = 1 or X = 2.
The only thing left is X = 0 which is the opposite of those cases above.
The opposite of "at least 1 black" is "zero black marbles", ie "2 white marbles".

The key take away is that because case B is false, we know that marble #2 and marble #3 are both white.

So far we've shown that marbles 2 through 4 must be white.

Case A says that "first and second marbles have at least one white marble"
We could have marble #1 to be white or it could be black.
Marble 2 is locked in to be white.

B = black
W = white

The scenarios could be WWWW or BWWW
The first scenario referring to all white, and the second scenario is where the first is black and the rest are white.

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Let's calculate the probability of WWWW

The marbles selected are not replaced.

m = P(1st is white) = 8/16
n = P(2nd is white) = 7/15
r = P(3rd is white) = 6/14
s = P(4th is white) = 5/13
We have the numerators and denominators with their own countdowns.
The countdowns are directly due to the fact the marbles are not replaced.

m*n*r*s = (8/16)*(7/15)*(6/14)*(5/13)
m*n*r*s = (8*7*6*5)/(16*15*14*13)
m*n*r*s = 1680/43680
m*n*r*s = 1/26

The probability of case WWWW happening is 1/26

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Now to the probability for the BWWW case.

m = P(1st is black) = 8/16
n = P(2nd is white) = 8/15
r = P(3rd is white) = 7/14
s = P(4th is white) = 6/13
We also have a countdown, but slightly different for the numerators.
The first marble being black means the white count does not go down until after the second selection is made.
The denominators decrease the same as before.

m*n*r*s = (8/16)*(8/15)*(7/14)*(6/13)
m*n*r*s = (8*8*7*6)/(16*15*14*13)
m*n*r*s = 2688/43680
m*n*r*s = 4/65

The probability of getting BWWW in that order is 4/65

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The two probabilities found in the previous two sections are then added up
P(WWWW) + P(BWWW) = 1/26 + 4/65
P(WWWW) + P(BWWW) = 5/130 + 8/130
P(WWWW) + P(BWWW) = (5+8)/130
P(WWWW) + P(BWWW) = 13/130
P(WWWW) + P(BWWW) = 1/10
We can add these probabilities because the events WWWW and BWWW are mutually exclusive.

Answer: 1/10

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