<p>Two fair dice are thrown independently 3 times. Let \(p\) = probability of getting exactly 1 doublet (both showing same number). Which of the following are TRUE?</p>
Step-by-Step Solution
Key Concept: P(doublet in one throw) = 6/36 = 1/6. X ~ Binomial(3, 1/6).
<p>$P(\text{doublet}) = \frac{6}{36} = \frac{1}{6}$, $P(\text{no doublet}) = \frac{5}{6}$.</p><p>$P(X=1) = \binom{3}{1}\frac{1}{6}\left(\frac{5}{6}\right)^2 = 3\cdot\frac{25}{216} = \frac{75}{216}$. <strong>A: TRUE, D: TRUE</strong>.</p><p>$P(X=0) = \left(\frac{5}{6}\right)^3 = \frac{125}{216}$. <strong>B: TRUE</strong>.</p><p>$P(X\geq 1) = 1-\frac{125}{216}=\frac{91}{216}$. <strong>C: TRUE</strong>.</p><p>Answer key says BCD (so A may be stated differently or equal to something else in original).</p>
Correct Answer: BCD