Question

Two dice are thrown simultaneously. If X denotes the number of sixes, find the expectation of X.

Answer 1

Answer:

P(0) =  25/36

P(1)  = 5/18

p(2)  = 1/36

Step-by-step explanation:

Total Event = 6* 6 = 36

Number of  Sixes can be 0 , 1  & 2

Two sixes only (6 , 6)  - One Event

one six (1 , 6) , (2 , 6) , (3 , 6) , ( 4 , 6) , ( 5 , 6)

          & (6 , 1) , ( 6, 2) , (6 , 3) , (6 , 4) , (6 , 5)

= 10 Events

0 sixes = 36 - 1 - 10 = 25

P(0) =  25/36

P(1) =  10/36  = 5/18

p(2)  = 1/36

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

Two dice are thrown simultaneously. If X denotes the number of sixes

TO DETERMINE

The expectation of X

CALCULATION

Here two dice are thrown simultaneously

So the total number of possible outcomes = 6× 6 = 36

Now total number of possible outcomes for the event when no six obtain = 36 - 11 = 25

Since the number of event points containing atleast one six are 11 precisely

(1,6),(2,6),(3,6),(4,6),(5,6),(6,6),(6,1),(6,2),(6,3),(6,4),(6,5)

Now total number of possible outcomes for the event when one six obtain = 10

Since the event points containing one six are

(1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5)

Now total number of possible outcomes for the event when two six obtain = 1

Since the event points containing two six are (6,6)

$\displaystyle \sf{ \: }P(X=0) = Probability \: of \: no \: six = \frac{25}{36}$

$\displaystyle \sf{ \: }P(X=1) = Probability \: of \: one \: six = \frac{10}{36}$

$\displaystyle \sf{ \: }P(X=2) = Probability \: of \: two \: six = \frac{1}{36}$

So the probability distribution is given by

$\sf{ X : } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 2$

$\displaystyle\sf{ P(X ): } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{25}{36} \: \: \: \: \: \: \: \: \: \: \: \: \frac{10}{36} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{1}{36}$

Hence the required Expectation

$\sf{\sum \: X \: . \: P(X)}$

$= \sf{0. P(0) + 1. \: P(1) + 2 \:. P(2) \: }$

$= \displaystyle \sf{ \bigg(0 \times \frac{25}{36} \bigg) + \bigg(1 \times \frac{10}{36} \bigg) + \bigg(2 \times \frac{1}{36} \bigg) \: }$

$= \displaystyle \sf{ \frac{12}{36} \: }$

$= \displaystyle \sf{ \frac{1}{3} \: }$

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