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Okay I do know the answer is

$$ \frac{6^5 - 5^5}{6^5} $$ which gives you the probability of at least one six.

But then how can you find the probability of at least "Two" sixes? I think this should be related to choosing from a set? I'm not sure but I'd like to find out.

Toby
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Roy Kesserwani
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2 Answers2

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One approach is to consider the complementary event.

If you compute the probability of exactly zero 6's and exactly one 6, you can then take the complement of the probability to get "two or more".

If $X$ is the number of 6's observed

$$P(X\geq 2) = 1 - P(X=0) - P(X=1)$$

So with this approach - a very common one with probability problems - you reduce the problem to two equality events, each of which you should find easier to deal with.

However, if you already did some work, and have $P(X\geq 1)$ you can save yourself some calculation, since that's already $1 - P(X=0)$.

More generally, such problems are treated by using the binomial distribution.

If you have a set of independent trials (such as die rolls) with a constant probability of one event of interest (such as a 6), and you count the number of events of interest (each event of interest is traditionally called "a success") in $n$ such trials, then the number of successes is a binomially distributed random variable.

If $X$ is the number of successes in $n$ independent trials with constant probability of success, $p$, we're interested in $P(X=k)$ for some integer $k$ in ${0, 1, ..., n}$

First, two simpler problems: if we had $k$ trials all succeed, that would have probability $p^k$. If we had $m$ trials all fail, that would have probability $(1-p)^m$. I presume you can already see how to do those.

Now, we have $k$ trials of $n$ succeed and $n-k$ fail. If it was the first $k$ that succeeded, then that would have probability $p^k (1-p)^{n-k}$. But the successes can come in any of the $n$ positions, so we have to count all the different ways we can get $k$ successes in $n$ trials, which is the same as the number of arrangements of $k$ "$S$" symbols and $n-k$ "$F$" symbols. This is just ${n}\choose{k}$.

Glen_b
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  • HOW DO YOU FIND the probability of at least 3 sixes? It would be nice if you can give me an answer and show me how you did it. – Roy Kesserwani Apr 22 '14 at 01:01
  • Your question sounds too much like a standard textbook problem (possibly even homework) for me to give an explicit numerical answer. You can compute it by adding $P(X=4)+P(X=5)+P(X=6)$, or following the above approach, by taking $P(X\geq 2)-P(X=2)$. – Glen_b Apr 22 '14 at 01:04
  • The individual probabilities can be evaluated pretty directly, or you can use the binomial distribution as indicated in the answer. – Glen_b Apr 22 '14 at 01:09
  • THE HOMEWORK QUESTION IS THE IMAGE THAT I HAVE POSTED ON THE POST. I JUST WANT TO GET A DEEPER UNDERSTANDING.CLEARLY, I AM NOT JUST TRYING TO GET AN ANSWER HERE. – Roy Kesserwani Apr 22 '14 at 01:12
  • Given you asked for both a numerical answer and the working, it's not at all clear that the next question on the homework isn't exactly what you asked for. The link to the binomial gives explicit numerical formulas. If you explain where the difficulty lies, perhaps, I could maybe explain that. It doesn't help to shout. – Glen_b Apr 22 '14 at 01:16
  • See extended explanation of binomial probabilities. – Glen_b Apr 22 '14 at 01:27
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if X ist the random variable for the amount of sixes, then $P(X\geq1)$ means, that at least one 6 has to be thrown. With the converse probability it is $P(X\geq1)=1-P(X=0)$

For P(X=k) you can use the binomial Distribution $n \choose k $$ \cdot p^k \cdot (1-p)^{n-k} $

And $P(X \geq 2)=1-P(X=0)-P(X=1)$

greetings,

calculus

callculus42
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