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

Hint: 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
\[n! = n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)\left( {n - 4} \right) \ldots \ldots .2.1\]