Two-fair-dice-are-rolled-once-Let-X-be-the-random-variable-representing-the-sum-of-the-numbers-that-show-up-on-the-two-dice-Find-X-

Question Number 176168 by nadovic last updated on 14/Sep/22

Two fair dice are rolled once . Let X

be the random variable representing

the sum of the numbers that show

up on the two dice . Find X .

Answered by aleks041103 last updated on 14/Sep/22

l e t x_{1} a n d x_{2} b e t h e n u m b e r s o n t h e t w o d i c e .

t h e n X = x_{1} + x_{2}

t h e n t h e p r o b a b i l i t y o f g i v e n v a l u e X i s :

P (X) = \underset{x_{1} = 1}{\sum^{6}} p (x_{1}, X - x_{1}) =

= \underset{x = 1}{\sum^{6}} p (x) p (X - x) = \frac{1}{36} \underset{x = 1}{\sum^{6}} {\begin{cases} 1, 1 ⩽ X - x ⩽ 6 \\ 0, e l s e \end{cases}

1 ⩽ X - x ⩽ 6 \Rightarrow 1 - X ⩽ - x ⩽ 6 - X

\Rightarrow X - 6 ⩽ x ⩽ X - 1

\Rightarrow P (X) = \frac{1}{36} ∣ ([X - 6, X - 1] \cap [1, 6] \cap N) ∣

o b v . P (X < 2) = 0 a n d P (X > 12) = 0

f o r X \in [2, 6] : X - 6 \notin N w h i l e X - 1 \in N a n d 6 > X - 1 > 0

\Rightarrow P (X \in [2, 6]) = \frac{1}{36} ∣ [1, X - 1] \cap [1, 6] \cap N ∣=

= \frac{1}{36} ∣ {1, 2, \dots, X - 1} ∣= \frac{X - 1}{36}

f o r X = 7 : P (7) = \frac{1}{36} ∣ {1, 2, \dots, 6} ∣= \frac{1}{6}

f o r X \in [8, 12] : [X - 6, X - 1] \cap [1, 6] = [X - 6, 6]

\Rightarrow P (X \in [8, 12]) = \frac{1}{36} ∣ {X - 6, X - 5, \dots, 6} ∣=

= \frac{1}{36} (13 - X)

\Rightarrow P (X) = {\begin{cases} 0, X \notin [2, 12] \cap N \\ \frac{X - 1}{36}, X \in [2, 7] \cap N \\ \frac{13 - X}{36}, X \in [8, 12] \cap N \end{cases}

Commented by Tawa11 last updated on 15/Sep/22

Great sir

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