NIMCET 2015 — Mathematics PYQ
NIMCET | Mathematics | 2015If a fair dice is rolled successively, then the probability that 1 appears in an even numbered throw is:
Choose the correct answer:
- A.
365
- B.
365
- C.
61
- D.
115
(Correct Answer)
115
Explanation
1. Define the Probabilities for a Single Roll
When rolling a fair dice:
-
Probability of getting a 1, P(S)=61
-
Probability of NOT getting a 1, P(F)=1−61=65
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)3P(S)
-
Case 3 (6th throw): Failure on 1st through 5th, Success on 6th ⟹P(F)5P(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)=(61)(65)=365
-
Common ratio (r) = P(F)2=(65)2=3625
4. Calculate the Sum
Using the formula for the sum of an infinite geometric series S∞=1−ra:
Correct Option: (d)
Explanation
1. Define the Probabilities for a Single Roll
When rolling a fair dice:
-
Probability of getting a 1, P(S)=61
-
Probability of NOT getting a 1, P(F)=1−61=65
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)3P(S)
-
Case 3 (6th throw): Failure on 1st through 5th, Success on 6th ⟹P(F)5P(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)=(61)(65)=365
-
Common ratio (r) = P(F)2=(65)2=3625
4. Calculate the Sum
Using the formula for the sum of an infinite geometric series S∞=1−ra:
Correct Option: (d)