Question

Four numbers are chosen at random (without replacement) from the set {1, 2, 3, ....., 20}.
Statement-1: The probability that the chosen numbers when arranged in some order will form an AP Is $\dfrac{1}{{85}}$.
Statement-2: If the four chosen numbers from an AP, then the set of all possible values of common difference is \[\left\{ { \pm 1,{\text{ }} \pm 2,{\text{ }} \pm 3,{\text{ }} \pm 4,{\text{ }} \pm 5} \right\}.\]
A. Statement-1 is true. Statement-2 is true; Statement-2 is the correct explanation for Statement-1
B. Statement-1 is true. Statement-2 is false
C. Statement-1 is false. Statement-2 is true
D. Statement-1 is true. Statement-2 is true: Statement-2 is not the correct explanation for Statement-1

Answer 1

Hint: Write all possible ways in which 4 numbers can be chosen between 1 to 20 and can be in A.P. to derive a number of favourable outcomes, then divide it with the total number of possible outcomes to get the probability.

Complete step-by-step answer:
We know that from 20 numbers we can choose 4 numbers in $^{20}{C_4}$ ways.
Now let’s make all the possible cases in which four numbers can be chosen between 1 to 20 and are in A.P.
When a=1 and d=1 we get, 1,2,3,….,17,18,19,20 so when d=1 we get 17 cases which will form an A.P. when 4 numbers are selected.
When a=1 and d=2 we get, 1,3,5,….,14,16,18,20 so when d=2 we get 14 cases which will form an A.P. when 4 numbers are selected.
When a=1 and d=3 we get, 1,4, 7,….,11,14,17,20 so when d=3 we get 11 cases which will form an A.P. when 4 numbers are selected.
When a=1 and d=4 we get, 1,5, 9, …, 17so when d=4 we get 8 cases which will form an A.P. when 4 numbers are selected.
When a=1 and d=5 we get 1,6,11,….,16 so when d=5 we get 5 cases which will form an A.P. when 4 numbers are selected.
When a=1 and d=6 we get 1,7,13, 19 so when d=6 we get 2 cases which will form an A.P. when 4 numbers are selected.
We can see that the total number of cases when d$ \in $[1,2,3,4,5,6] makes an A.P.
So, probability that four numbers picked at random are in A.P. = $\dfrac{{Sum{\text{ }}of{\text{ }}A.P.}}{{^{20}{C_4}}}$ (equation 1)
Now we know that sum of A.P. = ${S_n} = (a + l)\dfrac{n}{2}$
So, sum of A.P. = $\dfrac{{(17 + 2)}}{2} \times 6$ (equation 2)
Using values from equation 2 in equation 1
The value of $^{20}{C_4}$=\[\dfrac{{n!}}{{r!}} \times (n - r)!\]
              =\[\dfrac{{20!}}{{4!}} \times (20 - 4)!\]
We get, probability that four numbers picked at random are in A.P. = \[\dfrac{{\dfrac{{(17 + 2)}}{2} \times 6}}{{\dfrac{{20!}}{{4!}} \times (20 - 4)!}}\]= $\dfrac{1}{{85}}$
So, Statement 1 is true i.e. The probability that the chosen numbers when arranged in some order will form an AP Is $\dfrac{1}{{85}}$.
But statement 2 is false because in statement 1 we choose numbers randomly and we will arrange them according to selection.
So, option B is correct

Note: One might be confused as multiple concepts are involved here like we have to find the probability of the numbers which are in A.P hence one must be thorough with basics of both AP and Probability as well.The important here is to realize the no. of cases for different values of a and d without which we may not arrive to the conclusion.