Question

If a number n is chosen at random from the set\left\\{ {1,2,3.....,1000} \right\\}. Then, the probability that$\;n$ is a number that leaves a remainder $1$, when divided by $7$, is,
A. $\dfrac{{71}}{{500}}$
B. $\dfrac{{143}}{{1000}}$
C. $\dfrac{{72}}{{500}}$
D. $\dfrac{{71}}{{1000}}$

Answer 1

To find the probability we need to find the sample space first. The sample space is nothing but the number of values in whole. The total number of values is called sample space. Then find the value of the number that satisfies the condition. To find the probability we need to divide the numbers that satisfy the condition to the sample space.

Formula:
Sample space be $n(s)$.
The total values satisfy the equation $n(A)$.
$p(A) = \dfrac{{n(A)}}{{n(s)}}$

Complete step-by-step answer:
Given that the set is from \left\\{ {1,2,3.....,1000} \right\\}.
Let the sample space be set because sample space is the set from which the number is taken and check whether it satisfies the condition.
While counting the values in the set, it is $1000$.
Hence, the total number of sets is $1000$.
As we are considering the set as sample space, the total number of sample spaces is $1000$.
Now, we have to look at the condition.
As per the question given, the condition is the number that leaves a remainder $1$, when divided by $7$.
Let us consider divided by $7$as the first condition and remainder $1$as the second condition.
First, consider the condition which is divided by $7$,
We will get set as the multiples of seven but it will not satisfy the second condition.
To satisfy the second condition, if we add $1$to the multiples of $7$, we will get the remainder as $1$when it is divided by $7$.
First, let us find the multiples of $7$which falls below $1000$. They are

$$7,{\text{ }}14,{\text{ }}21,{\text{ }}28,{\text{ }}35,{\text{ }}42,{\text{ }}49,{\text{ }}56,{\text{ }}63,{\text{ }}70,{\text{ }}77, \\\ {\text{ }}84,{\text{ }}91,{\text{ }}98,{\text{ }}105,{\text{ }}112,{\text{ }}119,{\text{ }}126,{\text{ }}133,{\text{ }}140, \\\ {\text{ }}147,{\text{ }}154,{\text{ }}161,{\text{ }}168,{\text{ }}175,{\text{ }}182,{\text{ }}189,{\text{ }}196,{\text{ }} \\\ 203,{\text{ }}210,{\text{ }}217,{\text{ }}224,{\text{ }}231,{\text{ }}238,{\text{ }}245,{\text{ }}252, \\\ {\text{ }}259,{\text{ }}266,{\text{ }}273,{\text{ }}280,{\text{ }}287,{\text{ }}294,{\text{ }}301,{\text{ }}308, \\\ {\text{ }}315,{\text{ }}322,{\text{ }}329,{\text{ }}336,{\text{ }}343,{\text{ }}350,{\text{ }}357,{\text{ }}364, \\\ {\text{ }}371,{\text{ }}378,{\text{ }}385,{\text{ }}392,{\text{ }}399,{\text{ }}406,{\text{ }}413,{\text{ }}420, \\\ {\text{ }}427,{\text{ }}434,{\text{ }}441,{\text{ }}448,{\text{ }}455,{\text{ }}462,{\text{ }}469,{\text{ }}476, \\\ {\text{ }}483,{\text{ }}490,{\text{ }}497,{\text{ }}504,{\text{ }}511,{\text{ }}518,{\text{ }}525,{\text{ }}532, \\\ {\text{ }}539,{\text{ }}546,{\text{ }}553,{\text{ }}560,{\text{ }}567,{\text{ }}574,{\text{ }}581,{\text{ }}588, \\\ {\text{ }}595,{\text{ }}602,{\text{ }}609,{\text{ }}616,{\text{ }}623,{\text{ }}630,{\text{ }}637,{\text{ }}644, \\\ {\text{ }}651,{\text{ }}658,{\text{ }}665,{\text{ }}672,{\text{ }}679,{\text{ }}686,{\text{ }}693,{\text{ }}700, \\\ {\text{ }}707,{\text{ }}714,{\text{ }}721,{\text{ }}728,{\text{ }}735,{\text{ }}742,{\text{ }}749,{\text{ }}756, \\\ {\text{ }}763,{\text{ }}770,{\text{ }}777,{\text{ }}784,{\text{ }}791,{\text{ }}798,{\text{ }}805,{\text{ }}812, \\\ {\text{ }}819,{\text{ }}826,{\text{ }}833,{\text{ }}840,{\text{ }}847,{\text{ }}854,{\text{ }}861,{\text{ }}868, \\\ {\text{ }}875,{\text{ }}882,{\text{ }}889,{\text{ }}896,{\text{ }}903,{\text{ }}910,{\text{ }}917,{\text{ }}924, \\\ {\text{ }}931,{\text{ }}938,{\text{ }}945,{\text{ }}952,{\text{ }}959,{\text{ }}966,{\text{ }}973,{\text{ }}980, \\\ {\text{ }}987,{\text{ }}994 \\\$$

Total there is $143$.
As to satisfy the second condition, if we add $1$to the above values, we will get the same number of terms as the last number will be $995$.
So the number of values that satisfy the condition are $143$.
To find probability we know that
Sample space be $n(s)$.
The total values that satisfy the equation $n(A)$.
$p(A) = \dfrac{{n(A)}}{{n(s)}}$
By substituting $n(s) = 1000$and $n(A) = 143$, we get
$p(A) = \dfrac{{143}}{{1000}}$
The probability will be $\dfrac{{143}}{{1000}}$.

Note: Remember to find the sample space first be careful while selecting it because there would be differences. Careful while taking values that satisfy the equation. The formula for the finding probability is the number of values that satisfy the condition to the number of sample spaces.
Divide the probability to the maximum extent to get the correct final answer