what is the probability that at least one is blue?

Question

if you have 15 chips of which 9 are red, 2 blue, 4 green and just 3 are taken randomly without replacement.

Answer 1

It is best to calculate the prob that none are blue.
= 13/15 * 12/14 * 11/13 = 22/35

P(at least one is blue) = 13/35

Answer 2

The probability of is union of two events, there is 1 blue chip drawn or there are 2 chips drawn. These events are X and Y respectively. This can be written out as:

P[at least one blue chip is drawn] = X U Y

From De Morgan's law, it is shown that the simplest method is to find the probability of 1 - P[no blue chips]:

X U Y = ((X U Y)')' = (X' ∩ Y')'
(X' ∩ Y')' → P[ (X' ∩ Y')'] = 1 - P[ (X' ∩ Y')]

The easiest way to see the last part is to draw a Ven diagram, which shows that (X' ∩ Y') is the event that no bluechips are drawn (more explicitly, the intersection of 1 chip is not drawn and 2 chips are not drawn.)

so in conclusion:

P[X U Y] = 1 - P[(X' ∩ Y')]

The rest of the problem is simple. Write out the probability of the 3 picks not having blue chips, and take 1 minus the intersection of the 3 events:
a blue chip is not drawn on the first draw: A = 13/15
a blue chip is not drawn on the second draw: B = 12/14
a blue chip is not drawn on the third draw: C = 11/13

Because A, B, and C are not independent trials (the outcome of A will affect the outcome of the B and C, i.e, conditional probability exists amongst the 3 events) the probability of not drawing a blue chip decreases with each successive trial.

The answer is:

1 - (13/15 * 12/14 * 11/13) = 13/35

Answer 3

It's easier to calculate the probability that NONE are blue, and then subtract that value from 1.

You start with 15 chips, of which 13 are not blue. Assume you pick a non-blue one (13/15).

Then you have 14 chips of which 12 are not blue. Assume you pick a second non-blue one (12/14).

Then you have 13 chips of which 11 are not blue. You'd need to pick a third non-blue one (11/13).

Multiply those three together:
(13/15) * (12/14) * (11/13) = 22/35
... to get the probability of getting no blue chips.

The probability of getting at least one blue chip is 1-(22/35) = 13/35.

Answer 4

There are 15C3 = 15*14*13/(3*2) = 455 ways of selecting 3 chips out of 15.

How many ways are there of selecting 3 chips where at least 1 is blue?

There are:

13C1 * 2C2 + 13C2 * 2C1 since there are 13 non-blue chips
13 * 1 + 78*2 = 169 ways of selecting at least 1 blue chip.

So the probability is 169/455 = 13/35 ~ 37%

Answer 5

The chances that the first chip chosen is blue is 2/15.
After once chip is chosen, there are 14 left, so the chances that the second one chosen is blue goes up to 2/14 minus the 2/15 chance of having picked the first one, since it might have been a blue one, and no longer available to be chosen.
Likewise, in picking the third chip, there are now only 13 chips left, 2, 1 or 0 could be blue at this point.
So the chance of it being blue is 2/13 minus 2/14 minus 2/15.
OR 0.154 - 0.143 - 0.133 = 0.083 or 8.3 %

Answer 6

Calculate the chance of not getting any blue:
First pull 13/15, 2nd 12/14 third 11/13
13/15*12/14*11/13 = 0.628571429
Chance of getting some blue = 1 - chance of getting zero blue
Chance of getting some blue = 0.371428571

Answer 7

2/15 or 2 to 15

Answer 8

number of chances/number of total possiblitities
or
2/15 = 13.3%

Answer 9

Well, I did it the old-fashioned hard way (i.e. probability tree in Excel).

37.1%