A die is thrown three times. Getting a 3 or a | vedclass

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(B) Let $n = 3$ be the number of trials.Success is defined as getting a $3$ or a $6$. The probability of success in a single trial is $p = \frac{2}{6}

Vedclass · Accepted Answer

$\frac{7}{27}$

Vedclass · Answer

(B) Let $n = 3$ be the number of trials.Success is defined as getting a $3$ or a $6$. The probability of success in a single trial is $p = \frac{2}{6} = \frac{1}{3}$.The probability of failure is $q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}$.We need the probability of at least two successes,which is $P(X \ge 2) = P(X = 2) + P(X = 3)$.Using the binomial distribution formula $P(X = k) = {}^nC_k \cdot p^k \cdot q^{n-k}$:$P(X = 2) = {}^3C_2 \cdot \left(\frac{1}{3}\right)^2 \cdot \left(\frac{2}{3}\right)^1 = 3 \cdot \frac{1}{9} \cdot \frac{2}{3} = \frac{6}{27} = \frac{2}{9}$.$P(X = 3) = {}^3C_3 \cdot \left(\frac{1}{3}\right)^3 \cdot \left(\frac{2}{3}\right)^0 = 1 \cdot \frac{1}{27} \cdot 1 = \frac{1}{27}$.Therefore,$P(X \ge 2) = \frac{6}{27} + \frac{1}{27} = \frac{7}{27}$.

$A$ die is thrown three times. Getting a $3$ or a $6$ is considered success. Then the probability of at least two successes is

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