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?

p_{1} > p_{2}

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p_{1} = p_{2} = 0.6651

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p_{1} < p_{2}

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p_{1} = p_{2} = 0.3349

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Solution

The correct option is A $p_{1} > p_{2}$
$p_{1} = 1 - (no die shows six)$
$p_{1} = 1 - (\frac{5}{6})^{6} = 0.665102$

$p_{2} = 1 - (no die show six, and one die shows six)$

$p_{2} = 1 - (\frac{5}{6})^{12} - \frac{12!}{1! \times 11!} (\frac{5}{6})^{11} \times \frac{1}{6} = 0.618667$

$p_{1} > p_{2}$

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Consider the following events.E1: Six fair dice are rolled and at least one die shows six.E2: Twelve fair dice are rolled and at least two dice show six.Let p1 be the probability of E1 and p2 be the probability of E2. Which of the following is true?

The correct option is A p1>p2p1=1−(no die shows six)p1=1−(56)6=0.665102p2=1−(no die show six, and one die shows six)p2=1−(56)12−12!1!×11!(56)11×16=0.618667p1>p2

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?

The correct option is A $p_{1} > p_{2}$
$p_{1} = 1 - (no die shows six)$
$p_{1} = 1 - (\frac{5}{6})^{6} = 0.665102$

$p_{2} = 1 - (no die show six, and one die shows six)$

$p_{2} = 1 - (\frac{5}{6})^{12} - \frac{12!}{1! \times 11!} (\frac{5}{6})^{11} \times \frac{1}{6} = 0.618667$

$p_{1} > p_{2}$