for "at least" problems, why do you need to "1-(none)"?

[eric beans]

for "at least" problems, why do you need to "1-(none)"?

 

[pka]

for "at least" problems, why do you need to "1-(none)"?

Pardon the double negative but at least one means not none.
So It is not the case that there is not the absent of all of some of a particular thing.
Example: An urn contains ten colored balls, four blue & six reds. Three are drawn at random.
What is the probability that at least one of the three is blue?
Well the probability that all are red is \(\dfrac{6}{10}\).
So \(1-\dfrac{6}{10}\) is the probability that not all red or at least one blue.

 

[JeffM]

for "at least" problems, why do you need to "1-(none)"?

You don’t. That is an error.

The correct formulation is probability of at least ONE = 1 - probability of none.

The probability of at least two = 1 minus the sum of the probabilities of exactly none and exactly one.

Get it now?

 

[eric beans]

Pardon the double negative but at least one means not none.
So It is not the case that there is not the absent of all of some of a particular thing.
Example: An urn contains ten colored balls, four blue & six reds. Three are drawn at random.
What is the probability that at least one of the three is blue?
Well the probability that all are red is \(\dfrac{6}{10}\).
So \(1-\dfrac{6}{10}\) is the probability that not all red or at least one blue.

i sort of get it but i still don't get it.

shouldn't there be a number that indicates how many picks? in your example, how come 3 picks doesn't factor into "at least one" equation?

i did a simple example to try to understand this. supposing out of 2 picks, what's the probability that at least one of them is a blue? instead of 1-6/10 =4/10, i got 52/90. is this right? for all instances of picking at least 1 blue out of picks up to 4 (limit of blue balls available) would the probability be 4/10 (i.e. = 1-6/10)? what am i doing wrong?

 

[eric beans]

You don’t. That is an error.

The correct formulation is probability of at least ONE = 1 - probability of none.

The probability of at least two = 1 minus the sum of the probabilities of exactly none and exactly one.

Get it now?

i'm trying to understand the logic here but i'm missing something and can't articulate it.

so say out of 4 red, 6 blue balls in an urn, what's the probability of picking out at least TWO red, it would be 1-(6/10+4/10) =0?

 

[JeffM]

Don’t go switching problems. You had 6 red, 4 blue. Pick two without replacement.

What are the probabilities

No blues = 1/3 = 5/15

One blue = 8/15

Two blues = 2/15

These are mutually exclusive and exhaustive so the probabilities should add to 1.

(5 + 8 + 2)/15 = 1. They do.

So I can calculate at least one blue in two ways.

8/15 + 2/15 = 10/15 = 2/3.

1 - 1/3 = 2/3.

I get the same answer because 1 - 1/3 = (1/3 + 8/15 + 2/15) - 1/3 = 8/15 + 2/15.

Here it makes no difference in terms of number of computations, but if you were talking about picking 5 balls and asking what is the probability of at least one blue ball, it is quicker to calculate the probability of no blue balls and subtract that from 1.

What were you doing with that stuff in green?

 

[Harry_the_cat]

Pardon the double negative but at least one means not none.
So It is not the case that there is not the absent of all of some of a particular thing.
Example: An urn contains ten colored balls, four blue & six reds. Three are drawn at random.
What is the probability that at least one of the three is blue?
Well the probability that all are red is \(\dfrac{6}{10}\).
So \(1-\dfrac{6}{10}\) is the probability that not all red or at least one blue.

In pka's example, you are drawing three balls.
You are asked for P(at least one is blue) which is equal to P(exactly 1 blue)+P(exactly 2 blue)+P(exactly 3 blue).
Now, you could work out all of those if you wanted to and add them up.
BUT
You also know that P(0 blue) +P(exactly 1 blue)+P(exactly 2 blue)+P(exactly 3 blue)=1, because they are all the outcomes.
SO
P(exactly 1 blue)+P(exactly 2 blue)+P(exactly 3 blue) = 1 - P(0 blue) THIS IS THE BIT I THINK YOU ARE NOT GETTING
This way, you only have to work out one probability rather than three.

 

[lex]

@eric beans
You are correct in your post #4
and your method (green writing) in post #4 is correct, but you calculated incorrectly: [math]\tfrac{\boldsymbol{24}}{90}+\tfrac{\boldsymbol{24}}{90}+\tfrac{12}{90}...[/math]
The event "at least one blue" = any outcome except both red (*)

So you can calculate the probability the way you said, adding the probability of each successful outcome:
[math]\tfrac{4}{10} \times \tfrac{6}{9} + \tfrac{6}{10} \times \tfrac{4}{9} + \tfrac{4}{10} \times \tfrac{3}{9} =\tfrac{60}{90} \hspace2ex \left(=\tfrac{2}{3}\right)[/math]
OR
you can use the statement (*) above, and work out the probability this way:
[math] \text{Prob of at least one blue = Prob(any outcome) - Prob(both red)}[/math][math]\hspace20ex = \hspace8ex 1\hspace12ex - \hspace2ex\tfrac{6}{10} \times \tfrac{5}{9}\\ = 1 \hspace2ex - \hspace2ex \tfrac{30}{90}\\ = \tfrac{60}{90} \hspace2ex \left(=\tfrac{2}{3}\right)[/math]


If you want to do @pka's question in post#2

The event 'at least one blue' = any outcome except all three red (*)

So, P(Event 'at least one blue')=

OR
using (*) above:
[math]\text{P(Event at least one blue) = P(any outcome) - P(all three red)}[/math][math]\hspace28ex =\hspace6ex 1 \hspace6ex - \hspace2ex \tfrac{6}{10} \times \tfrac{5}{9} \times \tfrac{4}{8} = 1 - \tfrac{120}{720} = \tfrac{5}{6}[/math]

 

[JeffM]

I am going to try again.

First, I figured out what you were trying to do in green. I think that may explain your confusion. (6/10) * (4/9) = 24/90, not 20/90. With arithmetic mistakes like that, you are going to have trouble understanding what is going on. Moreover, you did not compute one of the probabilities at all. Let’s calculate the probabilities correctly.

24/90 (first blue, then red)
24/90 (first red, then blue)
30/90 (both red)
12/90 (both blue)

Now if you add those up, you get (24 + 24 + 30 + 12)/90 = 90/90 = 1.

It will always be true that the sum of the probabilities of an exhaustive set of mutually exclusive events is 1. You can use it as a non-definitive check on your work.

Now let’s determine which probabilities include at least one blue.

They are (24/90) + (24/90) + (12/90) = 60/90 = 2/3.

What is the probability of no blue? 30/90 = 1/3. And obviously it is true that 1 - (1/3) = 2/3.

I suggest that you play around with more examples, but take care to get the arithmetic right. Different number of draws. Different number of at leasts. With replacement and without replacement. I suspect half an hour of careful numerical experimentation will make all clear.

 

[Steven G]

Suppose the random variable x can take on the values 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9.

Now p(0) + p(1) + p(2) + p(3) + p(4) + p(5) + p(6) + p(7) + p(8) + p(9) = 1

Suppose you want p(x>=5) = p(5) + p(6) + p(7) + p(8) + p(9)

p(0) + p(1) + p(2) + p(3) + p(4) + p(5) + p(6) + p(7) + p(8) + p(9) = 1

p(5) + p(6) + p(7) + p(8) + p(9) = 1 - [p(0) + p(1) + p(2) + p(3) + p(4)]