Question

A bag contains cards bearing numbers from 11 to 30. A card is taken out of the bag at random. Find the probability that the selected card has a multiple of 5 on it.

Answer 1

Hint: At first we’ll find the number of ways of selection any 1 card out of the total number of cards, then we’ll find the number of multiples of 5 from 11 to 30 and the number of ways of selecting one card with multiple of 5 from total cards having multiple of 5.
Then using the formula for probability we’ll find the required probability.

Complete step-by-step answer:
Given data: A bag contains cards bearing numbers from 11 to 30.
Total number of cards in the bag
 $ = 30 - 11 + 1$
 \[ = 20\]
We know that the number of ways of selecting any ‘r’ elements out of a total ‘n’ number of elements irrespective of the order of ‘r’ elements is given by ${}^n{C_r}$,
Where, ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ and $n! = n(n - 1)(n - 2)(n - 3)(n - 4)..........(3)(2)(1)$
Therefore the number of ways that any 1 card is chosen out of $20 = {}^{20}{C_1}$
Multiples of 5 from 11 to 30 are 15, 20, 25 and 30
Therefore, the number of cards giving a multiple of 5 $ = 4$
The number of ways that any 1 card having multiple of 5 is chosen out of 4 cards $ = {}^4{C_1}$
Therefore, the probability that the card is chosen is having a multiple of 5 $ = \dfrac{{{}^4{C_1}}}{{{}^{20}{C_1}}}$
Using ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
 $ = \dfrac{{\dfrac{{4!}}{{1!\left( {4 - 1} \right)!}}}}{{\dfrac{{20!}}{{1!\left( {20 - 1} \right)!}}}}$
Now using \[n! = n\left( {n - 1} \right)!\] , we get,
 $ = \dfrac{{\dfrac{{4 \times 3!}}{{1!\left( 3 \right)!}}}}{{\dfrac{{20 \times 19!}}{{1!\left( {19} \right)!}}}}$
On simplification we get,
 $ = \dfrac{4}{{20}}$
Dividing numerator and denominator by 4 we get,
 $ = \dfrac{1}{5}$
Hence, the probability that the selected card has a multiple of 5 on it is $\dfrac{1}{5}$.

Note: An alternative method for this solution can be
Number of cards having a multiple of 5 $ = 4$
Total number of cards $ = 20$
Therefore, the probability that the card is chosen is having a multiple of 5 $ = \dfrac{4}{{20}}$
Dividing numerator and denominator by 4
Therefore, the required probability $ = \dfrac{1}{5}$