A die is loaded in such a way that each odd number is twice as likely to occur as each even number. Find P(G), where G is the event that a number greater than 3 occurs on a single roll of the die.
A die is loaded in such a way that each odd number is twice as likely to occur as each even number. Find P(G), where G is the event that a number greater than 3 occurs on a single roll of the die.

Solution:

The probability of odd numbers
\begin{array}{l} =2 \times \text { (Probability of even number) } \\ \Rightarrow P(\text { Odd })=2 \times P \text { (Even) } \\ \text { Now, } P(\text { Odd })+P(\text { Even })=1 \\ \Rightarrow 2 P \text { (Even })+P(\text { Even })=1 \\ \Rightarrow 3 P \text { (Even) }=1 \\ P(\text { Even })=1 / 3 \end{array}
Therefore,
\mathrm{P}(\mathrm{O} \mathrm{dd})=1-\frac{1}{3}=\frac{3-1}{3}=\frac{2}{3}
The total no. occurs on a single roll of die =6
And the no. greater than 3=4,5 or 6
Therefore, P(G)=P (no. greater than 3)
=P( no. is 4,5 or 6)
So here, 4 and 6 are the even numbers and 5 is odd
\begin{array}{l} \therefore \mathrm{P}(\mathrm{G})=2 \times \mathrm{P}(\text { Even }) \times \mathrm{P}(\text { Odd }) \\ =2 \times 1 / 3 \times 2 / 3 \\ =4 / 9 \end{array}
As a result, the required probability is 4 / 9