If x denotes the number of sixes in four cons | vedclass

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(A) Let $n = 4$ be the number of trials (throws of a dice).Let $p$ be the probability of getting a 'six' in a single throw,so $p = \frac{1}{6}$.Let $q

Vedclass · Accepted Answer

$\frac{1}{1296}$

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(A) Let $n = 4$ be the number of trials (throws of a dice).Let $p$ be the probability of getting a 'six' in a single throw,so $p = \frac{1}{6}$.Let $q$ be the probability of not getting a 'six',so $q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}$.Since the throws are independent,this follows a binomial distribution $B(n, p)$.The probability of getting $x$ successes is given by $P(x = k) = {}^nC_k \cdot p^k \cdot q^{n-k}$.For $x = 4$,we have:$P(x = 4) = {}^4C_4 \cdot (\frac{1}{6})^4 \cdot (\frac{5}{6})^{4-4}$$P(x = 4) = 1 \cdot (\frac{1}{6})^4 \cdot (\frac{5}{6})^0$$P(x = 4) = 1 \cdot \frac{1}{1296} \cdot 1 = \frac{1}{1296}$.

If $x$ denotes the number of sixes in four consecutive throws of a dice,then $P(x = 4)$ is

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