Question

Cards bearing numbers $1,3,5, \ldots 35$ are kept in a bag. A card is drawn at random from the bag. Find the probability of getting a card bearing
(i) a prime number less than $15$.
(ii) a number divisible by $3$ and $5$.

Answer 1

Hint: Here, we have to find the probability of getting a card bearing a prime number less than $15$ and a number divisible by $3$and $5$. We will use the probability formula which is defined as the ratio of the number of favourable outcomes to the total number of favourable outcomes. Firstly, we will calculate the number of favourable outcomes of a number less than $15$ and the number divisible by $3$ and $5$. The Favourable outcomes for numbers which are less than $15$ are the numbers which can only be divisible by \[1\] and the number itself and the favourable outcomes for a number divisible by $3$and $5$ are the numbers which are divided by both $3$and $5$.

Complete step-by-step answer:
Probability can be defined as the mathematical term for the likelihood that something will occur. It is the ability to understand and estimate the possibility of a different combination of outcomes.
It is the ratio of number of favourable outcomes to the total number of favourable outcomes. i.e.,
$P(E) = \dfrac{{No.\,\,of\,\,favourable\,\,outcomes}}{{Total\,\,no.\,\,of\,\,favourable\,\,outcomes}}$
As given cards are marked with numbers from $1$ to $35$.
Therefore, total number of outcomes $ = 35$

(i) Let ${E_1} = $ Event of getting a prime number less than $15$
Numbers which are prime and less than $15$ are $3,5,7,11,13$
Number of favourable outcomes ${E_1} = 5$
Total number of outcomes $ = 35$
Using the probability formula. We get,
$ \Rightarrow P({E_1}) = \dfrac{5}{{35}} = \dfrac{1}{7}$
Hence, the probability required to get a prime number less than $15$ is $\dfrac{1}{7}$.

(ii) Let ${E_2} = $Event of getting a number which is divisible by $3$ and $5$
 Numbers which are divisible by $3$ and $5$ $ = 15,\,30$
 Number of favourable outcomes $ = 2$
 Using the probability formula. We get,
$ \Rightarrow P({E_2}) = \dfrac{2}{{35}}$
Hence, the probability required to get a number which is divisible by $3$ and $5$ is $\dfrac{2}{{35}}$

Note: One should carefully write the number of favourable outcomes in each case as forgetting some number can make the whole calculation wrong for finding the probability and remember that the total number of outcomes remain the same in each case. The probability of an event is always greater or equal to zero but can never be less than zero. If S is the sample space then the probability of occurrence of sample space is always $1$ . That means if the experiment is performed then it is sure to get one of the sample spaces.