ME Test Series

Question

Two dice are thrown simultaneously. The expected sum of the numbers shown up is?

Comments

7.

Minimum sum is 2, maximum sum is 12. Sum them up and take average.

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7 ???

Answer 1

Possible outcomes$= X = \left \{ 2,3,4,5,6,7,8,9,10,11,12 \right \}$

$P(2) =\left \{ 1,1 \right \} \large = \frac{1}{36}$

$P(3) =\left \{ 1,2 \right \}, \left \{ 2,1 \right \} = \large \frac{2}{36}$

$P(4) =\left \{ 1,3 \right \}, \left \{ 2,2 \right \} ,  \left \{ 3,1 \right \}  = \frac{3}{36} $

$P(5) = \large \frac{4}{36} $

$P(6) = \large \frac{5}{36} $

$P(7) =  \large \frac{6}{36} $

$P(8) = \large \frac{5}{36} $

$P(9) = \large \frac{4}{36} $

$P(10) = \large \frac{3}{36}$

$P(11) = \large \frac{2}{36}$

$P(12) = \large \frac{1}{36}$

 

$E[X] = \sum_{X=2}^{12} X \times P_x = \large \frac{4+6+12+20+30+42+40+36+30+22+12}{36} = \frac{252}{36} = 7$

Comments

kumar.dilip don't know probably work 

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@kumar.dilip, just try for the 3 dices.
If the distribution comes out to be symmetric, then the expected value will just be equal to the mid point., 

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@Mk Utkarsh

$P\left ( 2 \right )=\frac{2}{36}$

but u have written $P\left ( 2 \right )=\frac{4}{36}$ in place of summation.

Please edit it

Answer 2

if n fair dices are rolled simultanously then expected sum of numbers is 3.5n here dices are 2 there so we will have 3.5*2=7