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Let's roll $10$ dice and consider those cases when at least one $1$ appears. What is the probability than among these cases two or more $1$'s appear?

Solution.

We have to find a conditional probability. Let's define the events:

  • (a) $A$ is the event that at least one $1$ appears.
  • (b) $B$ is the event that at least two $1$'s appear.

The answer is $P(B\mid A)$, which indicates the probability of at least two coming out if it comes out at least one. From the definition of conditional probability we have

$$P(B\mid A) = \frac {P(B\cap A) }{ P(A)}.$$

Since every event containing at least two $1$'s contains at least one $1$; one has then that $B\cap A = B$. We have to calculate $P(B)$ and $P(A)$. So $$P(B\mid A) = \frac {P(B) }{ P(A)}.$$

For this we calculate $P(\overline A)$ and $P(\overline B)$. Where $\overline A$ denotes the event that no $1$ appears and $\overline B$ is the event in which at most one $1$ appears.

$$P\left(\overline A\right) = \frac{5^{10}} {6^{10}}$$ $$P\left(\overline B\right) = \frac{ 5^{10} + 5^9\times10 } {6^{10}}$$

And finally

$$P(A)=1-P\left(\overline A\right) \text{ and } P(B)=1-P\left(\overline B\right).$$

...

I don't understand the complementary probabilities. Can you explain this solution?

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    what is unclear? – 2017-01-21
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    what exactly is not clear? The complementary probabilities? – 2017-01-21
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    Same question: After my edits, what is unclear? – 2017-01-21
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    I don't understand the complementary probabilities. – 2017-01-21
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    Why don't you go for binomial distribution? – 2017-01-21
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    @Masacroso: So, the numbers come from **Bin[10,(1/6)]** then **P(A) = 1 - f(0)** and **P(B) = 1 - [ f(0) + f(1) ]**? – 2017-01-21
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    The distribution of $X$, in my answer, is a binomial, yes, over $10$ "positions" with probability $1/6$. – 2017-01-21
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    Ok, now I have the same in my notebook. Thanks you very much. – 2017-01-21

3 Answers 3

1

Instead of use $A$ and $B$ we will define these events in a different manner that IMO it is much easier to understand. We define the random variable $X$ that count the number of $1$'s in the throw.

By example, if we write $\Pr[X=2]$ this mean the probability that there are exactly two $1$'s in the throw. Then, to symbolize the probability of the event $A$ at least one $1$ we write

$$\Pr[A]=\Pr[X\ge 1]$$

To symbolize the probability of the event $B$ at least two $1$'s we write

$$\Pr[B]=\Pr[X\ge 2]$$

Now, observe that

$$\Pr[X\ge 1]=\sum_{k=1}^{10}\Pr[X=k]=1-\Pr[X=0]$$

In other words: the complementary to $\Pr[X\ge 1]$ is $\Pr[X=0]$. You can see this of this way too: the values of $X$ comes from the set $E=\{0,1,\ldots,10\}$, that is or zero $1$'s, or one $1$, or two, etc. up to $10$.

Then at least one $1$ is $A=\{1,2,\ldots,10\}$ and the complementary to $A$ is $E-A=\{0\}$, that is, zero $1$'s (no one).

For the other case we have that

$$\Pr[X\ge 2]=\sum_{k=2}^{10}\Pr[X=k]=1-\Pr[X=0]-\Pr[X=1]$$

The notation $A$ and $B$ hides the explicit relations. I hope that with this notation you can understand the solution of the exercise more easily.

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What is the complementary event that "at least one $1$ appears"? It is that "no $1$ appears", isn't it? The number of the cases when no $1$ appears is

$$P\left(\overline A\right) = \frac{5^{10}} {6^{10}}$$

If no $1$ appears then there are only $5$ remaining numbers. The number of possibilities to select $10$ times from the set $\{2,3,4,5,6\}$ is $5^{10}$. The total number of possibilities is, of course $6^{10}.$

As far as the event that "at least two $1$'s appear" is that "at most one $1$ appears."

There are two possibilities: "no $1$ appears" and "exactly one $1$ appears". The number of those case is

$$5^{10} + 5^9\times10 .$$

So

$$P\left(\overline B\right)=\frac{5^{10} + 5^9\times10}{6^{10}}.$$

0

Considering you have main problem here.

For this we calculate $P(\overline A)$ and $P(\overline B)$. Where $\overline A$ denotes the event that no $1$ appears and $\overline B$ is the event in which at most one $1$ appears.

$P\left(\overline A\right) = \frac{5^{10}} {6^{10}}$$ $$P\left(\overline B\right) = \frac{ 5^{10} + 5^9\times10 } {6^{10}}$

Explanation -

  1. $\overline A$ denotes the event that no $1$ appears.

Then we have probability of getting any other number except 1.

So probability = $\frac56$

And as 10 rolls.

Probability = $\left(\frac56\right)^{10}$

  1. $\overline B$ is the event in which at most one $1$ appears.

It means either 0 times one appears or 1 times one in 10 rolls.

So we have,

Probability = $\left(\frac56\right)^{10} + \left( \frac16\right) \cdot \left( \frac56\right)^9$

= $\frac{(5)^{10}}{(6)^{10}} + \frac{1 \cdot (5)^9}{(6)^{10}}$

= $\frac{(5)^{10} + 1 \cdot (5)^9}{(6)^{10}}$

So you have one mistake instead of 10 in numerator its 1.