Question:
Two dice are thrown successively 4 times and getting a doublet is called a success. The probability of getting at least 1 success is
Two dice are thrown successively 4 times and getting a doublet is called a success. The probability of getting at least 1 success is
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Use the complement: 1 minus the probability of no doublet in four throws.
Updated On: Oct 1, 2026
- \(\frac{50}{1296}\)
- \(\frac{1246}{1296}\)
- \(\frac{625}{1296}\)
- \(\frac{671}{1296}\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
A throw of two dice has 36 outcomes, and 6 of them are doublets. So \(p = \dfrac16\) and \(q = \dfrac56\) per throw. Four independent throws make a binomial experiment with \(n = 4\).
Step 2: Use the complement:
\[ P(\text{at least one}) = 1 - P(\text{none}) = 1 - \left(\frac56\right)^4 \]
Step 3: Compute:
\[ \left(\frac56\right)^4 = \frac{625}{1296} \]
\[ 1 - \frac{625}{1296} = \frac{671}{1296} \]
Step 4: Result:
Option (D). Option (C) \(\dfrac{625}{1296}\) is the probability of no success at all.
Final Answer:
The probability is 1 - (5/6)^4 = 671/1296. \[ \boxed{\text{(D) }\dfrac{671}{1296}} \]
A throw of two dice has 36 outcomes, and 6 of them are doublets. So \(p = \dfrac16\) and \(q = \dfrac56\) per throw. Four independent throws make a binomial experiment with \(n = 4\).
Step 2: Use the complement:
\[ P(\text{at least one}) = 1 - P(\text{none}) = 1 - \left(\frac56\right)^4 \]
Step 3: Compute:
\[ \left(\frac56\right)^4 = \frac{625}{1296} \]
\[ 1 - \frac{625}{1296} = \frac{671}{1296} \]
Step 4: Result:
Option (D). Option (C) \(\dfrac{625}{1296}\) is the probability of no success at all.
Final Answer:
The probability is 1 - (5/6)^4 = 671/1296. \[ \boxed{\text{(D) }\dfrac{671}{1296}} \]
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