Question

Three numbers are chosen from 1 to 30. The probability that they are not consecutive is
A) $\dfrac{{142}}{{145}}$
B) $\dfrac{{144}}{{145}}$
C) $\dfrac{{143}}{{145}}$
D) $\dfrac{1}{{145}}$

Answer 1

Hint: According to the question we have to find the probability that the three numbers are chosen from 1 to 30 are not consecutive so first of all we have to find the probability that the three numbers are chosen from 1 to 30 are consecutive.
So, to find the probability that the three numbers are chosen from 1 to 30 are consecutive first of all we have to find the total number of outcomes means the sample space as given in the questions we have to choose numbers from 1 to 30 so we have to find the sample space from all these 30 numbers. Now, as we know we have to choose 3 numbers to make a group of three consecutive numbers so we will make the group with the given numbers from 1 to 30.
Now, we have to consider the three numbers as one digit, so the numbers will be 28 and those three numbers can be arranged in three different ways. Hence, we can find the number of ways of three numbers that are consecutive.
Now, we have to find the probability that the three numbers are chosen from 1 to 30 consecutively by dividing the favourable event with the sample space or the total number of outcomes. But according to the question we have to find the probability that the three numbers are chosen from 1 to 30 not consecutive so we have to subtract the obtained probability when the numbers are consecutive by 1.

Complete step by step answer:
Step 1: As askes in the question we have to find the probability that the three numbers chosen from 1 to 30 are not consecutive so first of all we have to find the probability that the three numbers chosen from 1 to 30 are consecutive. So, now we have to find the total number of outcomes or the sample space from the given numbers 1 to 30.
Hence,
The sample space is $ = 30!$
Step 2: Now, in this step we have to make groups of three consecutive numbers from 1 to 30, which can be possible as $(1,2,3)$ to $(28,29,30)$ Hence, there will be 28 pairs.
Step 3: Now, we have to consider the three numbers as a single number/digit which can be arranged in the following ways given below:
Ways of arranging three numbers $3!$
Step 4: From step 3 now we can find our favourable outcomes by considering 3 numbers as single digit, the numbers will be 28 and those three numbers can be arranged in three different ways.
Favourable outcomes $28! \times 3!$
Step 5: Hence, the probability that the three numbers chosen from 1 to 30 are consecutive can be found by dividing our favourable outcomes by total outcomes or sample space.
Probability $ = \dfrac{{28! \times 3!}}{{30!}}$
On solving the obtained expression,
$ = \dfrac{{28! \times 3 \times 2 \times 1}}{{30 \times 29 \times 28!}}$
On eliminating $28!$ from the obtained expression,
$
   = \dfrac{6}{{30 \times 29}} \\
   = \dfrac{1}{{145}} \\
 $
Step 5: Now, we can find the required probability that the three numbers are chosen from 1 to 30 not consecutive so we have to subtract the obtained probability when the numbers are consecutive by 1.
Required probability$ = 1 - \dfrac{1}{{145}}$
On solving,
Required probability$ = \dfrac{{144}}{{145}}$
Final solution: Hence, on finding sample space or total outcomes and favourable events we have find the probability that the three numbers are chosen from 1 to 30 are not consecutive is $ = \dfrac{{144}}{{145}}$

Hence, the correct option is (B).

Note:
If the probability of an event to be occur is $p(E)$ then the probability of the event not to be occur is $p(\overline E ) = 1 - p(E)$
To solve the value of factorial we can use the formula $n! = n(n - 1)(n - 2)(n - 3)(n - 4)........2 \times 1$ where $n$ is the given number.
Sample is the number of total possible outcomes and favourable outcomes are the event which is/ are required.