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What is the probability of rolling exactly two sixes in $7$ rolls of a die?

I know this is a binomial probability.

$P(X=2)=\binom{7}{2}(1/6)^2(5/6)^5$.

By the definition of probability formula "Probability formula is the ratio of number of favorable outcomes to the total number of possible outcomes." Then why probability of rolling exactly two sixes in $7$ rolls of a die isn't $=2/42$? The denominator is $42$ because in one die there are $6$ faces, and in $7$ rolls of a die, there are $7\times 6=42$ faces.

user81411
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  • Because you have two dices I think and so "two ways" of having two sixes, even though it doesn't seem to make much sense – Euler_Salter Dec 29 '16 at 02:53
  • I don't quite understand your question. The "probability of rolling exactly two sixes in 6 rolls of a die" should be ${6\choose 2}{1 \over 6^2}{5 \over 6^4} \neq {2\over 42}$ no? – gowrath Dec 29 '16 at 03:10
  • I imagine you are asking why the probability of rolling exactly two sixes in 7 rolls of a die isn't $2 \over 42$? – gowrath Dec 29 '16 at 03:13
  • @gowrath yes, you're right. I am editing. – user81411 Dec 29 '16 at 03:23

2 Answers2

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The total number of possible outcomes is $6^7$. In each try we have $6$ possibilities, and we repeat it for $7$ times. So $6^7$ in denomerator.

The number of favorable outcomes is $\binom{7}{2}5^5$. To see why, consider one of the cases such as 66?????, where each ? can have $5$ possibilities. So for this specific selection of the order of the two six, we have $5^5$ possibilities. There are $\binom{7}{2}$ possibilities for selecting the positions of the two $6$.

The probability is therefore $$\frac{\binom{7}{2}5^5}{6^7}$$

msm
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  • Suppose I have 7 balls, among which 3 are red and 4 green. What is the probability that if I randomly draw 4 balls two of them are red? can I use here the formula $\text{Probability}=\frac{\text{number of favorable outcomes}}{\text{total number of possible outcomes}}$? I suspect because total number of possible outcome is not $2^4$, since if first 3 balls are red, the fourth ball must be green. – user81411 Dec 30 '16 at 09:02
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    @user81411 The number of favorable outcomes is $\binom{4}{2}\binom{3}{2}$ and the total number of outcomes is $\binom{7}{4}$. – msm Dec 30 '16 at 10:20
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    hm..., it becomes hypergeometric. Thank you very much. – user81411 Dec 30 '16 at 12:36
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Define $X_{1},X_{2},\ldots,X_{7}$ as Bernoulli random variables, such that $X_{i}=1$ when the $i$th die lands $6$ and $X_{i}=0$ otherwise. Let $X=\sum_{i=1}^{6}{X_{i}}$ be the number of sixes in $7$ die rolls.

Probability of $X$ sixes in $7$ rolls of a die.

The number of sixes $X$ follows binomial distribution. $X\sim{Binomial(7,1/6)}$.

$\displaystyle{P(X=2)={{7}\choose{2}}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^5}$

Probability of a certain outcome, given that you land $2$ sixes in $7$ rolls of a die.

The naive definition of probability which states that,

$$P(A)=\frac{\text{number of sample points in }A}{\text{total number of points in sample space }S}$$

is applicable, when the sample space $S$ is finite and the sample points are equally likely.

Given that we land $2$ sixes in $7$ die rolls, i.e. the event $\{X=2\}$, what is the likelihood of any particular outcome, say,

$\{X_{1}=0,X_{2}=1,X_{3}=0,X_{4}=1,X_{5}=0,X_{6}=0,X_{7}=0\}$

Each such outcome occurs with equal probability. The sample space $S$ consists of ${{7}\choose{2}}=21$ such sequences. Therefore,

$\displaystyle{P(X_{1}=0,X_{2}=1,X_{3}=0,X_{4}=1,X_{5}=0,X_{6}=0,X_{7}=0|X=2)=\frac{1}{21}}$

Given $X=2$, these are uniformly distributed.

Note. In the strict sense of term, I have computed conditional probability above. It is $P(A|B)=P(AB)/P(B)$, where $P(AB)=(1/6)^2(5/6)^5$ and $P(B)={{7}\choose{2}}(1/6)^2(5/6)^5$.

Quasar
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