Question

A die is loaded so that the probability of face i is proportional to i, $i=1,2,...6$ . The probability of an even number occurring when die is rolled $\dfrac{k}{7}$ . Find the value of k.

Answer 1

Hint: First, we are told that faces are proportional to numbers on die. So, we can write it as $1=k,2=2k,....6=6k$ . Then we will add all the probability and equate it with 1. We know that maximin probability is equal to 1 and minimum is equal to 0. After this we will get the value of k. Then, we are asked to find even numbers. On finding that, we will put the value of k in that and then compare it with $\dfrac{k}{7}$ to get the value of k.

Complete step-by-step answer:
Here, we are told that the faces on die are proportional to the numbers on them. So, we can write it as $1=k,2=2k,....6=6k$ .
Now, we know that probability is always between 0 and 1. So, we can write summations of all these equal to 1.
$\therefore k+2k+3k+4k+5k+6k=1$
On further solving, we get value of k as
$21k=1$
$k=\dfrac{1}{21}$ …………………….(1)
Now, here we are asked to find the probability of even numbers. So, we can write it as
$2k+4k+6k=12k$
Now, we will substitute the value of k from equation (1) in the above equation. So, we will get as
$12k=12\times \dfrac{1}{21}$
On further solving, we get as $\dfrac{4}{7}$ . Now, on comparing this with $\dfrac{k}{7}$ , we can say that the value of k is 4.
Thus, the value of k is 4.

Note: Remember here k is taken as probability constant and nothing else. We can use any other constant also. Also, students sometimes make mistakes in considering the value of k as $k=\dfrac{1}{21}$ and writing the answer which will be wrong. So, do not make this mistake and read the question carefully.