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Hint: The cases that can occur are either both dices do not have six on their faces, either one has or the last case is both dices have six on its faces.

Here, $X$ be the representation used for the number of sixes that are obtained when two dices are thrown simultaneously.

Therefore, $X$ can take either value as $0,1$or $2$.

$P\left( {X = 0} \right) = P$ (Not getting six on any of the dice) $ = \dfrac{{25}}{{36}}$

$P\left( {X = 1} \right) = P$(Six on first die and non- six on second die) $ + P$(Non- six on first die and six on second die)$ = 0.33$ $ = 2\left( {\dfrac{1}{6} \times \dfrac{5}{6}} \right) = \dfrac{{10}}{{36}}$

$P\left( {X = 2} \right) = P$(Six on both the dices) $ = $ $\dfrac{1}{{36}}$

Now, the next step is to find the expectation.

Therefore,

Expectation of $X = E\left( X \right) = \sum {{X_i}P\left( {{X_i}} \right)} $

$ = 0 \times \dfrac{{25}}{{36}} + 1 \times \dfrac{{10}}{{36}} + 2 \times \dfrac{1}{{36}}$

$ = \dfrac{1}{3} = 0.33$

Answer $ = 0.33$

Note: Do not forget to take the case where $X$ can have the value zero.

Here, $X$ be the representation used for the number of sixes that are obtained when two dices are thrown simultaneously.

Therefore, $X$ can take either value as $0,1$or $2$.

$P\left( {X = 0} \right) = P$ (Not getting six on any of the dice) $ = \dfrac{{25}}{{36}}$

$P\left( {X = 1} \right) = P$(Six on first die and non- six on second die) $ + P$(Non- six on first die and six on second die)$ = 0.33$ $ = 2\left( {\dfrac{1}{6} \times \dfrac{5}{6}} \right) = \dfrac{{10}}{{36}}$

$P\left( {X = 2} \right) = P$(Six on both the dices) $ = $ $\dfrac{1}{{36}}$

Now, the next step is to find the expectation.

Therefore,

Expectation of $X = E\left( X \right) = \sum {{X_i}P\left( {{X_i}} \right)} $

$ = 0 \times \dfrac{{25}}{{36}} + 1 \times \dfrac{{10}}{{36}} + 2 \times \dfrac{1}{{36}}$

$ = \dfrac{1}{3} = 0.33$

Answer $ = 0.33$

Note: Do not forget to take the case where $X$ can have the value zero.

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