GATE2025_CS1_Q8

A fair six-faced dice, with the faces labelled '1', '2', '3', '4', '5', and '6', is rolled thrice. What is the probability of rolling ‘6' exactly once?

✅ Final Answer:

The correct answer is option (A) 75216.

🔹 Short Answer:

This is a binomial probability problem. The probability of rolling a '6' (success) is 1/6, and not rolling a '6' (failure) is 5/6. We want exactly one success in three rolls. The number of ways this can happen is C(3, 1) = 3. The total probability is the number of ways multiplied by the probability of one specific sequence: 3 × (1/6)¹ × (5/6)² = 3 × (1/6) × (25/36) = 75/216.

🔸 Long Answer:

This problem involves calculating the probability of a specific outcome in a series of independent trials. Since each dice roll is independent and has only two outcomes of interest (rolling a '6' or not rolling a '6'), this is a classic application of the Binomial Probability formula.

Step 1: Define Probabilities for a Single Trial

• The total number of outcomes on a single roll is {1, 2, 3, 4, 5, 6}.
• Let 'S' be the event of rolling a '6' (Success). The probability is P(S) = 1/6.
• Let 'F' be the event of not rolling a '6' (Failure). The probability is P(F) = 5/6.

Step 2: Identify Possible Sequences for the Event

We need to roll a '6' exactly once in three rolls. The possible sequences are:

1. Success on 1st roll: S, F, F (6, not 6, not 6)
2. Success on 2nd roll: F, S, F (not 6, 6, not 6)
3. Success on 3rd roll: F, F, S (not 6, not 6, 6)

There are 3 distinct ways this can happen. This can also be calculated using the combinations formula "n choose k", where n=3 trials and k=1 success: C(3, 1) = 3! / (1! * (3-1)!) = 3.

Step 3: Calculate the Probability

Since the events are mutually exclusive (they cannot happen at the same time), we can add their probabilities.

Probability of one sequence (e.g., S, F, F) = P(S) × P(F) × P(F) = (1/6) × (5/6) × (5/6) = 25/216.

Since there are 3 such sequences, each with the same probability:

Total Probability = (Prob of Seq 1) + (Prob of Seq 2) + (Prob of Seq 3)
Total Probability = (25/216) + (25/216) + (25/216)
Total Probability = 3 × (25/216)
Total Probability = 75/216

Bibliography:
- Sheldon, R. (2014). A First Course in Probability. Pearson. (Chapter on Discrete Random Variables).
- "Binomial Distribution." Khan Academy.

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