Question

Cards bearing numbers $2,4,6,8,10,12,14,16,18$ and $20$ are kept in a bag. A card is drawn a random from the bag. Find the probability of getting a card which is: a prime number, a number divisible by$4$, a number that is a multiple of $6$, an odd number.

Answer 1

Step 1:Use the formula of probability

$P\left(A\right)=\frac{\text{Number of outcome favorable to A}}{\text{Total number of Possible outcome}}$

or

$P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}$

where,

• $P\left(A\right)$ is probability of an event ‘$A$’
• $n\left(A\right)$ is the favorable outcome of the event ‘$A$’
• $n\left(S\right)$ is total number of possible outcomes in a sample space

Step 2: Find the total number of possible outcomes

Cards bearing numbers $2,4,6,8,10,12,16,18$ and $20$ are kept in a bag.

$S=\{2,4,6,8,10,12,14,16,18,20\}$

$S$ is a sample space of all the possible outcomes which is 10

Thus, total number of possible outcomes in a sample space is

$n\left(S\right)=10$

Step 3: Find the probability of getting a prime number

Here, among ten outcomes only one contains a prime number

$A=\left\{2\right\}$

Thus, $n\left(A\right)=1$

$\text{Required probability}=\frac{1}{10}$

Step 4: Find the probability of getting a divisible by 4

Here, among $10$ outcomes, only five numbers are divisible by 4

$A=\{4,8,12,16,20\}$

Thus, $n\left(A\right)=5$

$\text{Rrequired probability}=\frac{5}{10}=\frac{1}{2}$

Step 5: Find the probability of getting a multiple of $6$

Here, among$10$ outcomes only three numbers are the multiple of $6$

$A=\{6,12,18\}$

Thus, $n\left(A\right)=3$

$\text{Required probability}=\frac{3}{10}$

Step 6: Find the probability of getting an odd number

Here, among$10$ outcomes all are even number.

So, $A=nil$

Thus, $n\left(A\right)=0$

$\text{Rrequired probability}=\frac{0}{10}=0$

Hence, the probability of getting a card which is a prime number is $\frac{1}{10}$, a number divisible by 4 is $\frac{1}{2}$, a number that is a multiple of 6 is $\frac{3}{10}$, an odd number is $0$.