NIMCET 2015 — Mathematics PYQ

When rolling a fair dice:

• Probability of getting a 1, $P(S)=\frac{1}{6}$

• Probability of NOT getting a 1, $P(F)=1-\frac{1}{6}=\frac{5}{6}$

2. Identify the Favorable Cases

We want the outcome "1" to appear for the first time on an even-numbered throw (2nd, 4th, 6th, 8th, ...).

• Case 1 (2nd throw): Failure on 1st, Success on 2nd $⟹P(F)P(S)$

• Case 2 (4th throw): Failure on 1st, 2nd, 3rd, Success on 4th $⟹P(F{)}^{3}P(S)$

• Case 3 (6th throw): Failure on 1st through 5th, Success on 6th $⟹P(F{)}^{5}P(S)$

  And so on.

3. Form the Infinite Series

Total Probability ($P$) is the sum of these mutually exclusive cases:

This is an infinite geometric series with:

• First term ($a$) = $P(S)P(F)=(\frac{1}{6})(\frac{5}{6})=\frac{5}{36}$

• Common ratio ($r$) = $P(F{)}^2=(\frac{5}{6}{)}^2=\frac{25}{36}$

4. Calculate the Sum

Using the formula for the sum of an infinite geometric series $S_{\infty }=\frac{a}{1-r}$:

Correct Option: (d)