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?

Engineering

Mathematics

Addition Theorem on Probability

Basic Definitions Related to Probability

Binomial Probability Theorem

Question

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?

$P 1 = P 2 = 0.3349$

$P 1 < P 2$

$P 1 > P 2$

$P 1 = P 2 = 0.6651$

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P₁ = 1 – (no die shows six)

$p_{1} = 1 - (\frac{5}{6})^{6} = 0.665102$

P₂ = 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