$100$ dices are thrown.How many are expected to fall either a $3$ or $6?$
Now expectation means, sum of values of dice , divided by number of possible outcome.
So, I have done like, $(3+6)/100$
But answer given like , it is nothing but the mean value , i.e. $100*(1/6+1/6)$
Which one correct to find expectation?
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$\begingroup$Consider the random variables $$X_1,X_2,...,X_{100}$$where $X_i=1$ if the roll results in $3$ or $6$ and $X_i=0$ otherwise. Now, note that $$P(X_i=1)=\frac{1}{3}=1-P(X_i=0)$$ and $$\mathbb{E}(X_i)=\frac{1}{3} +0 =\frac{1}{3}$$So the total number of rolls resulting in either a $3$ or a $6$ is nothing but $$\sum_{i=1}^n X_i$$So, expected no of dice resulting in $3$ or $6$ is: $$\mathbb{E}(\sum_{i=1}^n X_i)=\sum_{i=1}^n \mathbb{E}(X_i) = 100×\frac{1}{3}=\frac{100}{3}$$
$\endgroup$ 6 $\begingroup$Expectation is the mean/average value of something. In this case, they ask for the number of dice (out of 100) that would land on 3 or 6 (on average). Note that the values of the dice don't matter (the value of each side of the die could be 101 to 106 for all we care and the answer would be the same). What matters is that each die has 6 sides and each side is equally likely. On each roll, each die has a 1/6 chance of landing on a 3 and a 1/6 chance of landing on a 6. So if you roll 1 die, it has a 2/6 chance of landing on either a 3 or a 6. If you roll N dice, (2/6)*N would land on either a 3 or 6. So if you roll 100 dice, (2/6)*100 would land on a 3 or a 6. Note that the question asks for the number of dice landing on certain values and not for the average value of the sides that it lands on which would give a different answer and is a somewhat in the direction that your first answer was trying to go (but it still isn't right).
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