Key to "If you roll 5 standard six-sided dice, what’s..."

persepsi1ea

Answered question

2022-02-14

If you roll 5 standard six-sided dice, what’s the probability that you get at least two 4s?

Answer & Explanation

Amubbemibren3k

Beginner2022-02-15Added 10 answers

Let X denote number of die that show face '4' in $n =$ independent rolls of die, where each has a probability $p = 0.166667$ .
We denot this by:
$X \sim B i n o m i a l (n = 5, p = 0.166667)$
Probability of x number of die that show face '4' is:
$P (X = x) = (\begin{matrix} n \\ x \end{matrix}) p^{x} (1 - p)^{(n - x)}$
where $(\begin{matrix} n \\ x \end{matrix})$ is combinatorial expression:
$(\begin{matrix} n \\ x \end{matrix}) = \frac{n!}{x! (n - x)!}$
$P (X \geq 2), n = 5, p = 0.166667$
$P (X \geq 2) = 1 - P (X < 2)$
We first calculate:
$P (X < 2) = P (0 \leq X < 2) = P (0 \leq X \leq 1) = P (X = 0, 1)$
$= P (X = 0) + P (X = 1)$
$P (X = 0) = (\begin{matrix} 5 \\ 0 \end{matrix}) (0.166667)^{0} (1 - 0.166667)^{5} = 1 (0.166667)^{0} (0.833333)^{5} \approx 0.401878$
$P (X = 1) = (\begin{matrix} 5 \\ 1 \end{matrix}) (0.166667)^{1} (1 - 0.166667)^{4} = 5 (0.166667)^{1} (0.833333)^{4} \approx 0.401878$
$P (X = 0) + P (X = 1)$
$= 0.401878 + 0.401878 = 0.803756 \approx 0.804$
$P (X < 2) = 0.804$
$P (X \geq 2) = 1 - P (X < 2) = 1 - 0.804 = 0.196$
$P (X \geq 2) = 0.196$

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