Question

If a number x is chosen at random from the numbers 2,3,5,7 and the number y is chosen at random from the number 3,4,6,8, then P(xy= multiple of 3)?
a. $\dfrac{4}{7}$
b. $\dfrac{16}{11}$
c. $\dfrac{5}{8}$
d. $\dfrac{5}{16}$

Answer 1

Hint: First we find all the numbers in x and y. When 3 is present in x then the product of xy for each and every value of y are multiple of 3. Similarly, the multiple of 3 present in y in that case, the product of xy for each and every value of x are also multiple of 3. Using this logic, we will find the probability of the event.

Complete step-by-step answer:
First we find the total number of outcomes or total number of ways to choose x and y.
So, there are 4 values in x and similarly 4 values in y.

The total number of ways to choose x and y $=4\times 4=16$.

For xy to be multiple of 3 either multiple of 3 should be present in x or multiple of 3 should be present in y.

When the value of x is 3, in that case 3,4,6 and 8 are corresponding values of y.

In y, when the value of y = 3, in that case 2,3,5 and 7 are the corresponding value of x.
For y =6, the value of x can be 2,3,5,7.

They all form the favorable outcomes space of the event.
Total number of favorable outcomes $=4+3+3$.
The total number of favorable outcomes $=10$.

So, the probability of the event which given in the question is,
P (xy = multiple of 3)$=\dfrac{10}{16}$
P (xy = multiple of 3)$=\dfrac{5}{8}$

The probability that the product of x and y is multiple of 3 $=\dfrac{5}{8}$.

Therefore, option (c) is correct.

Note: The key step in solving this problem is the knowledge of sample space and favorable outcome for the given event. If we specify both correctly, then we are required to solve mathematical expressions for obtaining the answer. Hence, for solving probability problems we must proceed by forming sample space and favorable outcomes.