Question

Three integers are selected simultaneously from the set of integers $\\{1,2,3,...,50\\}$. The probability that the selected numbers are consecutive, is equal to
(A) $\dfrac{9}{\left( 25 \right)\left( 49 \right)}$
(B) $\dfrac{6}{\left( 25 \right)\left( 49 \right)}$
(C) $\dfrac{3}{\left( 25 \right)\left( 49 \right)}$
(D) $\dfrac{1}{\left( 25 \right)\left( 49 \right)}$

Answer 1

We are given a question based on probability and we are asked to find the probability of selecting three consecutive numbers from the given set. Three consecutive numbers refers to the set like, $[1,2,3],[2,3,4],...,[48,49,50]$, that is, we can have a total of 48 such sets. The number of ways three numbers can be selected from the set of 50 numbers is by using the combination formula, that is, $^{50}{{C}_{3}}$. In order to find the required probability, we will use the formula, $Probability=\dfrac{No.\text{ }of\text{ }favorable\text{ }outcomes}{Total\text{ }No.\text{ }of\text{ }outcomes}$ and then we will calculate the ratio of the same. Hence, we will have the value of the probability.

Complete step by step answer:
According to the given question, we are given a question asking us to find the probability of selecting three consecutive numbers from the given set of numbers.
The set of numbers given to us is,
$\\{1,2,3,...,50\\}$
Consecutive numbers refer to continuous numbers without missing out any number in between. For example - $[1,2,3],[2,3,4],...,[48,49,50]$
In the given set, we can have 48 such set of three consecutive numbers.
From the given set of numbers, number of ways in which three numbers can be picked up simultaneously can be found by using the combinations rule, which is, $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
So, we have the number of ways of selecting 3 numbers from a set of 50 numbers as,
$^{50}{{C}_{3}}$
We get the simplified form as,
$^{50}{{C}_{3}}=\dfrac{50!}{3!\left( 50-3 \right)!}$
${{\Rightarrow }^{50}}{{C}_{3}}=\dfrac{50!}{3!47!}$
Opening up the factorial in the numerator such that the factorial in the denominator get cancelled. We have,
${{\Rightarrow }^{50}}{{C}_{3}}=\dfrac{50\times 49\times 48\times 47!}{3!47!}$
${{\Rightarrow }^{50}}{{C}_{3}}=\dfrac{50\times 49\times 48}{3!}$
That is, we have,
${{\Rightarrow }^{50}}{{C}_{3}}=\dfrac{50\times 49\times 48}{6}$
Now, we will find the probability of the selecting three consecutive numbers, using the formula of probability, which is,
$Probability=\dfrac{No.\text{ }of\text{ }favorable\text{ }outcomes}{Total\text{ }No.\text{ }of\text{ }outcomes}$
Now, substituting the values, we get,
$\Rightarrow Probability=\dfrac{48}{^{50}{{C}_{3}}}$
$\Rightarrow Probability=\dfrac{48}{\dfrac{50\times 49\times 48}{6}}$
Rearranging the above expression, we get,
$\Rightarrow Probability=\dfrac{48\times 6}{50\times 49\times 48}$
Now, cancelling out the common terms, we get,
$\Rightarrow Probability=\dfrac{6}{50\times 49}$
Simplifying further, we get,
$\Rightarrow Probability=\dfrac{3}{25\times 49}$

So, the correct answer is “Option C”.

Note: In the above solution, we made use of the combinations rule and not permutation. Permutation is used when a particular has to be followed, but in the above solution where we had to find the total number of ways three numbers can be selected, we used the combinations rule. Do not get confused between which to use. Also, the probability formula should be written correctly and substitution should be done in a correct manner.