## Conditional Probability of 3 dice

• 公開日時: 2018/11/01 13:42
• 閲覧数: 137
• コメント数: 4
• カテゴリ: 入試・教育

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1
Strange to say, I have received the answers from four persons, but each answer is all different. I wonder, whose answer is Correct.

2
おじゃま虫です。

m ≦ A,B,C ≦ 6
$N _{m, 6}\ =\ (7-m) ^{3}$　とおり。

min{A,B,C} = m
$6 ^{3} p(E _{2})\ =\ N _{m, 6}\ -\ N _{m+1, 6}\ =\ (7-m) ^{3}\ -\ (6-m) ^{3}$　とおり。

m ≦ A,B,C ≦ M
$N_{m, M}\ =\ (M-m+1) ^{3}$　とおり。

min{A,B,C} = m，　Max{A,B,C} = M，
$6 ^{3} p(E _{1} \cap E _{2})\ =\ N_{m, M}\ -\ N_{m+1, M}\ -\ N_{m, M-1}\ +\ N_{m+1, M-1}$
$=\ (M-m+1) ^{3}\ -\ 2 (M-m) ^{3}\ +\ (M-m-1) ^{3}$
$=\ 6 (M-m)$　とおり。

$6 ^{3} p(E _{2})\ =\ 5 ^{3}\ -\ 4 ^{3}\ =\ 61$　とおり。
$6 ^{3} p(E _{1} \cap E _{2})\ =\ 6 (4-2)\ = 12$　とおり。

$p(E _{1}\ |\ E _{2})\ =\ \frac{p(E _{1} \cap E _{2})}{p(E _{2})}\ =\ \frac{12}{61},$　……　答

12とおりの出目 (A,B,C) は
(2,2,4) (2,4,2) (4,2,2)
(2,3,4) (2,4,3) (3,2,4) (3,4,2) (4,2,3) (4,3,2)
(2,4,4) (4,2,4) (4,4,2)

prime_132 さん 2018/11/02 17:11:48 報告
3
Formidable !

My answer was same as yours, prime_132 ^^

4
Solution from USA

Let $E_1$ denote the event $\max(A,B,C) = 4$ and $E_2$ denote the event $\min(A,B,C) = 2$. We have

$p(E_1 \mid E_2) = \frac{p(E_1 \cap E_2)}{p(E_2)}$

$p(E_2)$ is fairly easy to compute. There are $5^3 - 4^3 = 61$ possible outcomes with minimum value 2 ($5^3$ outcomes where each number is at least 2, subtract $4^3$ outcomes where each number is at least 3). To find $p(E_1 \cap E_2)$, each of $A$, $B$, $C$ must be between 2 and 4 inclusive, and at least one value must be 2, at least one must be 4. Then the outcomes are 234 in some order (6 outcomes), 224 in some order (3 outcomes) and 244 in some order (3 outcomes) giving 12 outcomes. The probability is $\frac{12}{61}$.