A die is formed so that the probability of getting a number $i$ when it is rolled is proportional to $i$.…
A die is formed so that the probability of getting a number $i$ when it is rolled is proportional to $i$. $(i=1,2,3,4,5,6)$. The probability of getting an odd number on the die when it is rolled is
$\frac{1}{2}$
$\frac{4}{7}$
$\frac{2}{7}$
$\frac{3}{7}$
Solution
Let the probability of getting a number $i$ is $P(i)$.
So,
$
\begin{aligned}
& P(i)=K i \\
& \Rightarrow \quad P(1)=K, P(2)=2 K, \quad P(3)=3 K \text {, } \\
& P(4)=4 K, \quad P(5)=5 K \text {, } \\
& P(6)=6 K \\
& \Sigma P(i)=1 \Rightarrow 21 K=1 \Rightarrow K=\frac{1}{21} \\
&
\end{aligned}
$
Now, probability of getting an odd number
$
=P(1)+P(3)+P(5)=9 K=\frac{9}{21}=\frac{3}{7} .
$