Question

Cards are numbered from 1 to 25. Two cards are drawn one after the other. Find the probability that the number on one card is a multiple of 7 and on the other is a multiple of 11.

Answer 1

Hint: To find the probability we list out the multiples of 7 and 11 respectively and we use the probability formula, probability of an event =$\dfrac{{{\text{number of favorable outcomes}}}}{{{\text{total number of outcomes}}}}$.

Complete step-by-step answer:
Given Data, cards are numbered from 1 to 25.
The cards which are multiples of 7, (from 1 – 25) are 7, 14 and 21. And the cards which are multiples of 11 are 11 and 22.
We have to find the probability that the number on one card is a multiple of 7 and on the other is a multiple of 11, i.e.
We have to pick one number from {7, 14, 21) and one from {11, 22}.
The probability of favorable outcomes of a given set is given by
P = ${}^{\text{n}}{{\text{C}}_{\text{r}}} = \dfrac{{{\text{n!}}}}{{{\text{r!}}\left( {{\text{n - r}}} \right)!}},{\text{ (where the symbol ! denotes factorial)}}$
Where n is the total elements in the set and r is the number of elements we want to pick.
Probability of picking multiple of 7 = ${}^3{{\text{C}}_1}$
And probability of picking a multiple of 11 = ${}^2{{\text{C}}_1}$
And the total number of possibilities of picking two numbers from numbers 1 to 25 = ${}^{25}{{\text{C}}_2}$
Hence the required probability = $\dfrac{{{\text{number of favorable outcomes}}}}{{{\text{total number of outcomes}}}}$= $\dfrac{{{}^3{{\text{C}}_1} \times {}^2{{\text{C}}_1}}}{{{}^{25}{{\text{C}}_2}}}$
$\Rightarrow {\text{Probabiliity P = }}\dfrac{{\dfrac{{{\text{3!}}}}{{{\text{1!}}\left( {{\text{3 - 1}}} \right)!}} \times \dfrac{{{\text{2!}}}}{{{\text{1!}}\left( {{\text{2 - 1}}} \right)!}}}}{{\dfrac{{{\text{25!}}}}{{{\text{2!}}\left( {{\text{25 - 2}}} \right)!}}}} \\\ \Rightarrow {\text{P = }}\dfrac{{\dfrac{{{\text{3!}}}}{{2!}} \times \dfrac{{{\text{2!}}}}{{{\text{1!}}}}}}{{\dfrac{{{\text{25!}}}}{{{\text{2!}}\left( {{\text{23}}} \right)!}}}} = \dfrac{{3 \times 2}}{{\left( {\dfrac{{25 \times 24}}{2}} \right)}} = \dfrac{1}{{50}} \\\$ --- (n! of a number is = n(n-1)(n-2)(n-3)……….) and 0! = 1.

Therefore the probability of drawing two cards from 1 to 25 where one is a multiple of 7 and other is a multiple of 11 is$\dfrac{1}{{50}}$.

Note: In order to solve questions of this type the key is to know the formulae of probability and the probability of a number of favorable outcomes from a given set. It is also important to understand the simplification of P =${}^{\text{n}}{{\text{C}}_{\text{r}}} = \dfrac{{{\text{n!}}}}{{{\text{r!}}\left( {{\text{n - r}}} \right)!}}$, using the concept of factorials.
The factorial of a positive integer n, denoted by n! is the product of all positive integers less than or equal to n.