Consider the following events:E_1: Six fair d | vedclass

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(A) For event $E_1$,six fair dice are rolled. The probability that no die shows a six is $(\frac{5}{6})^6$. Thus,$p_1 = 1 - (\frac{5}{6})^6 = 1 - 0.33

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$p_1 > p_2$

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(A) For event $E_1$,six fair dice are rolled. The probability that no die shows a six is $(\frac{5}{6})^6$. Thus,$p_1 = 1 - (\frac{5}{6})^6 = 1 - 0.3349 = 0.6651$.For event $E_2$,twelve fair dice are rolled. Let $X$ be the number of dice showing a six. $X$ follows a binomial distribution $B(n=12, p=\frac{1}{6})$.$p_2 = P(X \geq 2) = 1 - [P(X=0) + P(X=1)]$.$P(X=0) = (\frac{5}{6})^{12} \approx 0.1122$.$P(X=1) = \binom{12}{1} (\frac{1}{6})^1 (\frac{5}{6})^{11} = 12 	imes \frac{1}{6} 	imes 0.1346 \approx 0.2692$.$p_2 = 1 - (0.1122 + 0.2692) = 1 - 0.3814 = 0.6186$.Since $0.6651 > 0.6186$,we have $p_1 > p_2$.

Consider the following events:
$E_1$: Six fair dice are rolled and at least one die shows six.
$E_2$: Twelve fair dice are rolled and at least two dice show six.
Let $p_1$ be the probability of $E_1$ and $p_2$ be the probability of $E_2$. Which of the following is true?

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Consider the following events:$E_1$: Six fair dice are rolled and at least one die shows six.$E_2$: Twelve fair dice are rolled and at least two dice show six.Let $p_1$ be the probability of $E_1$ and $p_2$ be the probability of $E_2$. Which of the following is true?