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

• A. \(\frac{50}{1296}\)
• B. \(\frac{1246}{1296}\)
• C. \(\frac{625}{1296}\)
• D. \(\frac{671}{1296}\)

Hint:

Use the complement: 1 minus the probability of no doublet in four throws.

Answer 1

The Correct Option is D

Step 1: Binomial route:
$P(X \ge 1)$ with $X\sim B(4, \tfrac16)$.

Step 2: Sum the terms:
$P(X=1) = 4\cdot\tfrac16\cdot\tfrac{125}{216} = \tfrac{500}{1296}$. $P(X=2) = 6\cdot\tfrac1{36}\cdot\tfrac{25}{36} = \tfrac{150}{1296}$. $P(X=3) = 4\cdot\tfrac1{216}\cdot\tfrac56 = \tfrac{20}{1296}$. $P(X=4) = \tfrac1{1296}$.

Step 3: Add:
$500 + 150 + 20 + 1 = 671$, so the probability is $\tfrac{671}{1296}$, option (D).

Final Answer:
The probability is 671/1296. \[ \boxed{\text{(D) }\dfrac{671}{1296}} \]