Question

Question: What is the probability of these events when we randomly select a permutation of $\mathbf{\{}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{3}\mathbf{\}}$?

a)$\mathbf{1}$precedes$\mathbf{3}$.

b)$\mathbf{3}$precedes$\mathbf{1}$.

c)$\mathbf{3}$precedes$\mathbf{1}$and$\mathbf{3}$precedes$\mathbf{2}$

Short answer

Answer

The probability of $1$precedes $3$is $0.5$.

The probability of $3$precedes $1$is$0.5$ .

The probability of$3$precedes$1$and$3$precedes$2$is$\frac{1}{3}$.

Step-by-step solution

Given information.

Given Permutation-: $\{1,2,3\}$

Definition of the probability and formula used

Probability is the measure of the likelihood of an event to occur. Events can't be predicted with certainty but can be expressed as to how likely it can occur using the idea of probability.

Formula used:

Permutation:

Combination:

Calculating the probability for part(a)

First, we have to consider all the possibilities. For a set of three elements, there are only six permutations:

Those are the six of them, all the probabilities will be something out of six.

I count three $(A,B,C)$where 1 comes before 3

Thus the conclusion is that the probability is 0.5

Calculating the probability for part(b)

First, we have to consider all the possibilities. For a set of three elements, there are only six permutations:

Those are the six of them, all the probabilities will be something out of six.

I count three (D,F,E) where 3 comes before 1.

Thus, the probability is 0.5.

Calculating the probability for part(c)

First, we have to consider all the possibilities. For a set of three elements, there are only six permutations:

Those are the six of them, all the probabilities will be something out of six.

I count two (E,F) where 3 comes before 2 and 1,

Thus, the probability is$\frac{1}{3}$