Question

A ticket is drawn at random from a bag containing tickets numbered from $1$ to $40$. Find the probability that the selected ticket has a number which is multiple of $5$.

Answer 1

Here we will find the total number of possible outcomes from the $40$ cards and find the multiples of $5$ between $1$ and $40$ and with the help of it we will find the wanted outcomes, then divide the wanted outcomes by total outcomes to find the required.

Formula used:
Probability is a type of ratio where we compare how many times an outcome can occur compared to all possible outcomes.
${\text{Probability = }}\dfrac{{{\text{The number of wanted outcomes}}}}{{{\text{The number of possible outcomes}}}}$

Complete step-by-step answer:
It is given that a bag containing tickets numbered from $1$ to $40$.
And also given that a ticket is drawn at random from the bag.
So the possible outcome of choosing one ticket from ticket numbered $1$ to $40$ is $^{40}{C_1}$. Which is nothing but the combination of $1$ card out of $40$ cards.
Now we have to count the number which is multiple of $5$ between $1$ and $40$.
So the set of number which is multiple of $5$ between $1$ to $40$ is \left\\{ {5,{\text{ }}10,{\text{ }}15,{\text{ }}20,{\text{ }}25,{\text{ }}30,{\text{ }}35,{\text{ }}40} \right\\}
This is found because we have to find the probability that the taken card is a multiple of $5$.
So the possible outcome of choosing one ticket from the bag has a number which is multiple of $5$ is $^8{C_1}$.
That is nothing but the combination of $1$ card out of $8$ cards.
The probability that the selected ticket has a number which is multiple of $5$ is $\dfrac{{^8{C_1}}}{{^{40}{C_1}}}$
Using the combination formula we get the probability as
$= \dfrac{{\dfrac{{8!}}{{1!7!}}}}{{\dfrac{{40!}}{{1!39!}}}}$
Let us simplify these factorials in the above equation to get the required answer,
$= \dfrac{{\dfrac{{8.7!}}{{7!}}}}{{\dfrac{{40.39!}}{{39!}}}}$
By cancelling the terms in both numerator and denominator we have
= \dfrac{8}{{40}}$$$$ = \dfrac{1}{5}
Hence, the probability that the selected ticket has a number which is multiple of $5$ is $\dfrac{1}{5}$.

Note: A combination is a grouping or subset of items.
For a combination,
$C\left( {n,r} \right){ = ^n}{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$
Where, factorial n is denoted by $n!$ and it is defined by
$n! = n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)\left( {n - 4} \right) \ldots \ldots .2.1$