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by Fractal » Sat Jul 30, 2011 4:13 am
1) What is the chance of getting heads at least once in two fair coin flips?

P(H1 or H2) = P(H1) + P(H2) - P(H1 and H2)

= 1/2 + 1/2 - (1/2 * 1/2)

= 3/4


2) What is the probability that, on three rolls of a single fair die, at least one of the rolls will be a six?

I want to solve this second question with the approach used in the first question. I know that the (1-x) probability trick would be better for this question...

I tried it like the following:

P(6_1 or 6_2 or 6_3) = P(6_1) and P(6_2) and P(6_3) - P(6_1 and 6_2 and 6_3) - P(6_1 and 6_2) - P(6_1 and 6_3) - P(6_2 and 6_3)


P(6_1) and P(6_2) and P(6_3) = 1/6 + 1/6 + 1/6 = 3/6 = 1/2

P(6_1 and 6_2 and 6_3) = (1/6)^3 = 1/216

P(6_1 and 6_2) = (1/6)^2 = 1/36

P(6_1 and 6_3) = (1/6)^2 = 1/36

P(6_2 and 6_3) = (1/6)^2 = 1/36


Therefore:

1/2 - 1/216 - 3*(1/36) = 89/216

This solution is wrong, since the right probability is not 89/216 but 91/216.

Can anybody help?

thx a lot
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by Frankenstein » Sat Jul 30, 2011 8:34 am
Hi,
The formula you have used is incorrect. The correct one is:
p(AuBuC) = p(A)+p(B)+p(C)-p(AnB)-p(BnC)-p(AnC)+p(AnBnC)
Now, you will get the correct answer.
Cheers!

Things are not what they appear to be... nor are they otherwise
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by tpr-becky » Sat Jul 30, 2011 8:18 pm
a better way to think of at least one probability quesitons is to deal with them from the opposite side. The total probablility of an event happening or not happening is always 1 (there is 100% chance that the light is either raining or not raining). You can use this to your advantage in at least one problems

There is a probability of 1 that you will get at least one 6 or you will get no sixes - therefore you can use the formula 1 - (probability of getting none of the desired outcome).

in this problem it would be 1 - (probability of never getting a 6). The probability of not getting a 6 on an individual role is 5/6 - if you roll three times it is 5/6(5/6)(5/6)

Thus it is 1 - 125/216 = 91/216.

In your coin problem it would be 1 - (1/2 * 1/2) = 3/4

this approach works on any at least one probability problem.
Becky
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by Fractal » Sun Jul 31, 2011 4:18 am
Frankenstein wrote:Hi,
The formula you have used is incorrect. The correct one is:
p(AuBuC) = p(A)+p(B)+p(C)-p(AnB)-p(BnC)-p(AnC)+p(AnBnC)
Now, you will get the correct answer.
Thank you!

But why do we have to add this term: p(AnBnC)

In this example...

1) What is the chance of getting heads at least once in two fair coin flips?

P(H1 or H2) = P(H1) + P(H2) - P(H1 and H2)

... you have also to subtract the analog term P(H1 and H2)
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by HeintzC2 » Wed Aug 10, 2011 6:21 am
Agreed with Becky. The easiest way to do any "at least one" problem is to find the probability that NONE happen. When you subtract this figure from 1 (100%) the remainder is the probability for any event in which at least one occurs. This is much easier than trying to imagine all of the possibilities in which "at least one" could occur.

For the 6 sided die, the probability that no rolls are a 6 is: 5/6*5/6*5/6.
therefore, the probability that at least one roll is a 6 is 1-(5/6*5/6*5/6) = 216/216 - 125/216
P(at least one 6) = 91/216.
Fractal wrote:
Frankenstein wrote:Hi,
The formula you have used is incorrect. The correct one is:
p(AuBuC) = p(A)+p(B)+p(C)-p(AnB)-p(BnC)-p(AnC)+p(AnBnC)
Now, you will get the correct answer.
Thank you!

But why do we have to add this term: p(AnBnC)

In this example...

1) What is the chance of getting heads at least once in two fair coin flips?

P(H1 or H2) = P(H1) + P(H2) - P(H1 and H2)

... you have also to subtract the analog term P(H1 and H2)
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