Algebraic step including finite sum and binomial coefficient

Question

Hello, I've stumbled into the following algebraic step in my combinatorics text book. Beside calculating it directly, I can't find a proper justification for it.

Someone's got an idea?

Answer 1

This is $$\frac45\sum_{k=1}^{10}k\binom{10}k\frac{4^{10-k}}{5^{10}}$$ But $$k\binom{10}k=k\frac{10!}{k!(10-k)!}=10\frac{9!}{(k-1)!(10-k)!} =10\binom{9}{k-1}.$$ Your sum is $$8\sum_{k=1}^{10}\binom{9}{k-1}\frac{4^{10-k}}{5^{10}} =\frac85\sum_{j=0}^{9}\binom{9}{j}\frac{4^{9-j}}{5^{9}}=\frac85 \frac{(1+4)^9}{5^9}=\frac85$$ (using binomial theorem).

Answer 2

The point is to recognize that the sum $\sum_{k=0}^{10} \binom{10}{k} \cdot 0.2^k \cdot 0.8^{10-k} \cdot k$ is the expected value calculation of the expected number of successes in 10 successive, independent Bernoulli trials with probability of success $p=0.2$. The value of this sum is commonly known to equal $np$ where $n$ is the number of trials, which equals $10 \cdot 0.2 = 2$ in this case.

One way to remember the $np$ formula is to let $X$ be a random variable denoting the number of successful trials and to let $X_i$ be an indicator R.V. for the success of trial $i$. Then $$X = \sum_{i =1}^nX_i \implies E[X] = E \left[ \sum_{i=1}^n X_I \right] = \sum_{i=1}^nE[X_i] = \sum_{i=1}^n p = np$$

Answer 3

$$=0.8\sum_{k=0}^{10} \binom {10}{k} (0.2)^k * (0.8)^{10-k}.k$$ $$=(0.8)*(0.2)\sum_{k=1}^{10} \binom {9}{k-1} * (0.2)^{k-1} * (0.8)^{10-k}.k.\frac{10}{k}$$ $$=(0.8)*(0.2)*(10)\sum_{k=1}^{10} \binom {9}{k-1} (0.2)^{k-1} * (0.8)^{10-k}$$ Also we have

$$\sum_{k=1}^{10} \binom {9}{k-1} (0.2)^{k-1} * (0.8)^{10-k}=(0.8+0.2)^9=1^9=1$$

Hence

$$(0.8)*(0.2)*10\sum_{k=1}^{10} \binom {9}{k-1} (0.2)^{k-1} * (0.8)^{10-k}=(0.8)*(0.2)*(10)=1.6$$