Probability ,Settheory.

Question

$1)$ A random selector can only select one of the nine integers $\left \{ 1,2,.....,9 \right \}$ and it makes these selections with equal probability,Assume that  $n$ numbers are chosen this way.Find the probability such that the product of $n$ numbers selected is not divisible by $3$.

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$2)$ Let S = {1, 2, . . . , n} then Find |{(R, T) : R, T ⊆ S, R ∩ T = ∅}| = ? 

Comments

yes, What abt 2nd one ??

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we need ordered pair of (R,T) or unordered ???

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if it is ordered answer is $3^n$ ..i am writing a solution

Answer 1

First part answer in the above comments.

Answer to 2nd part:

Assuming, $\large\color{maroon}{\text{S}}$ = $\left \{ 1,2,3 \right \}$ , then following is the Relation in matrix representation :

$$\begin{align*} \text{Z} = \left \{ \left ( R,T \right ) \;\; | \; \; \text{R,T} \subseteq \; \text{S ,} R \cap T = \phi \right \} \end{align*}$$

example :

• first green circle means $\left \{ 2 \right \}$ and $\left \{ 3 \right \}$ is one pair $\left ( \left \{ 2 \right \},\left \{3  \right \} \right )$ in the relation $Z$. And because, we are considering ordered pairs ,then $\left ( \left \{ 3 \right \},\left \{2  \right \} \right )$ will also be a pair in relation $Z$.
• Second green circle means $\left \{ 2,3 \right \}$ and $\left \{ 1 \right \}$ is one pair $\left ( \left \{ 2,3 \right \},\left \{1  \right \} \right )$ in the relation $Z$. And we are considering ordered pairs then $\left ( \left \{ 1 \right \},\left \{2,3  \right \} \right )$ will also be a pair in relation $Z$.

Now,

No of such ordered pairs in relation $Z$ =  No of $1$'s in the matrix representation.

Counting procedure:

• one example: 2nd row of matrix, which is related to set $\left \{ 1 \right \}$.
• No of subsets that $\left \{ 1 \right \}$ will be related to are the subsets of $\left \{ 2,3 \right \}$.
• How many subsets of $\left \{ 2,3 \right \}$ ? . = $2^{3-1}$ .(because $1$ is missing here).
• Similarly for subsets $\left \{ 2 \right \}$ and $\left \{ 3 \right \}$ ,, i.e. 3rd and 4th row of the matrix.
• So total $1$'s for these $\binom{3}{1} = 3$ ,  one element subsets , is = $\binom{3}{1}*2^{3-1}$. (Or, total no of 1's in 2nd,3rd,4th rows of the matrix)

Doing in the same way, will end up arriving in the follwoing series,

$$\begin{align*} &=\binom{3}{0}*2^{3-0} + \binom{3}{1}*2^{3-1} + \binom{3}{2}*2^{3-2} + \binom{3}{3}*2^{3-3} \\ &=\binom{3}{0}*1^0*2^{3-0} + \binom{3}{1}*1^1*2^{3-1} + \binom{3}{2}*1^2*2^{3-2} + \binom{3}{3}*1^3*2^{3-3} \\ &=\sum_{r=0}^{3}\binom{3}{r}1^{r}.2^{3-r} \\ &= (1+2)^{3} \\ &=3^3 \end{align*}$$ 

If we assume $\large\color{maroon}{\text{S}}$ = $\left \{ 1,2,3,4,5,6....n \right \}$

• Then this above value will be $\large\color{red}{3^n}$

NOTE: 

• This relation Z is symmetric with only one self loop at $\phi$..
• Whenever we need only unordered pairs $(R,T) \;\; \text{see qs statement})$ then, we will only count off-diagonal $1$'s and one $(\phi,\phi)$ pair. So, how many unordered ?

 $$\begin{align*} \frac{3^n -1}{2} +1 \end{align*}$$

Comments

$3^n$ are total ordered pairs based on relation $Z$ constraint that includes pairs with $\phi$ also.

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@Debashish

What an explanation Buddy. But isn't it very simple to explain.

each element is having three choices.

A) to go in R

B) to go in T

C) don't go in either of them.

So 3^n.

Something wrong with it?

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No wrong with it. It is correct and elegant. Given QS based on relation, So I proceed with relation definition. You can add that as an answer.

Answer 2

each element is having three choices.

A) to go in R

B) to go in T

C) don't go in either of them.

So 3^n.