NIMCET 2008 — Mathematics PYQ

When two dice are rolled, the total number of possible outcomes is $6\times 6=36$.

• Event A (Sum is 5): The outcomes are $\{(1,4),(2,3),(3,2),(4,1)\}$.

  $$P(A)=\frac{4}{36}$$

• Event B (Sum is 7): The outcomes are $\{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\}$.

  $$P(B)=\frac{6}{36}$$

• Event C (Sum is neither 5 nor 7):

  $$P(C)=1-(P(A)+P(B))=1-\left(\frac{4}{36}+\frac{6}{36}\right)=1-\frac{10}{36}=\frac{26}{36}$$

2. Understanding "5 comes before 7"

The process stops as soon as a sum of 5 or 7 is rolled. We want the probability that the "stopping" roll is a 5. This can happen in several ways:

• 5 is rolled on the 1st attempt.

• Neither 5 nor 7 is rolled on the 1st attempt, but 5 is rolled on the 2nd attempt.

• Neither 5 nor 7 is rolled on the 1st and 2nd attempts, but 5 is rolled on the 3rd, and so on.

3. Mathematical Calculation (Infinite Series)

Let $P$ be the required probability:

This is an infinite geometric progression (GP) with first term $a=P(A)$ and common ratio $r=P(C)$.

Substituting the values: