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Roll a die $n$ times and let $M$ be the number of times you roll $6$. Assume that $n$ is large.

(a) Compute the expectation $EM$.
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EM means?
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srestha Expected Mean

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among 6 numbers '6' is getting once.

So, in 1 throw probability of getting 6 is $\frac{1}{6}$

in 2 throw  "               "       "          6 is $\frac{2}{6^{2}}$

in n throws     "                   "     "        6 is $\frac{n}{6^{n}}$

M is number of times 6 occurs on the dice.

Hence Probability that out of n throws m times 6 occurs will be :

Choose m throws out of n throws with nCm ways where 6 comes and then 6 comes with probability 1/6 in those m places and remaining places 6 does not come with 5/6 probability.

So, P {M = m} will be given by the below expression :

P{M = m} = nCm . (1/6)m . (5/6)(n-m)

This is a Binomial Distribution with p = 1/6.

Hence, EM = m=0 to n  m. nCm . (1/6)m . (5/6)(n-m)

Basically, EM = sum from m=0 to m=n of ( m . P{M = m})

We already know that this summation will simplify to np, the derivation of Expectation of a Binomial Distribution can be looked up in any standard book.

So, EM = np = n/6.

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