Let $p_i$ be the probability that a randomly chosen point has $i$ many friends, $i=0,1,2,3,4$. Let $X$ be a random variable such that for $i=0,1,2,3,4$, the probability $P(X=i)=p_i$. Then the value of $7E(X)$ is

Let pᵢ be the probability that a randomly chosen point has i many friends, i=0,1,2,3,4. Let X be a random variable such that for i=0,1,2,3,4, the probability P(X=i)=pᵢ. Then the value of 7E(X) is

$p_i = P(\text{a random point has } i \text{ friends}), \quad i = 0,1,2,3,4$

$X: P(X=i) = p_i, \quad i = 0,1,2,3,4$

$\text{To find: } 7E(X)$

$\text{Network: a line means the two points are friends}$ $C: 4 \text{ friends}$ $\text{Each of the other 6 points: } 3 \text{ friends}$ $1 \text{ point with } 4, \quad 6 \text{ points with } 3, \quad \text{total } 7 \text{ points}$ $p_4 = \dfrac17, \quad p_3 = \dfrac67, \quad p_0 = p_1 = p_2 = 0$ $\sum_i p_i = \dfrac17 + \dfrac67 = 1 \ \checkmark$ $E(X) = \sum_{i=0}^{4} i\,p_i$ $= 0\,p_0 + 1\,p_1 + 2\,p_2 + 3\,p_3 + 4\,p_4$ $E(X) = 3\cdot\dfrac67 + 4\cdot\dfrac17 = \dfrac{18+4}{7} = \dfrac{22}{7}$ $E(X) = \dfrac{\text{total degree}}{\text{total points}} = \dfrac{6\times3 + 1\times4}{7} = \dfrac{22}{7}$ $7E(X) = 7 \times \dfrac{22}{7} = 22$