John is blindfolded and shoots the ball towards the goal. You can assume that each time he shoots there is a probability of 0.2 to score and that shots are independent of each other. Every time he hits the target, John receives 2 points, and everytime he misses he loses 1 point. Compute the probability that after the four shots John will have more than two points .
My approach:
The Sample space of getting more than two points are the following ways
Let $S$ be the event of scoring and Let $S'$ be the event of missing
${SSSS,SSSS',SSS'S,SS'SS,S'SSS}$
Independently calculated the probability of each event
$P(SSSS)=1/625$
$P(SSSS')=4/625$
$P(SSS'S)=4/625$
$P(SS'SS)=4/625$
$P(S'SSS)=4/625$
Hence the probability of getting more than 2 points is
$P(X>2)=17/625$
Is this correct way to get the solution and i know its the topic binomial distribution but can't figure out a way to draw the table
He is only shooting $(4)$ times, and his point total will be greater than $(2)$ if and only if he hits the target either all $(4)$ times, or $(3)$ out of $(4)$ times.
That is, hitting the target $(2)$ out of $(4)$ times results in a point total of $(2 \times 2) - (2 \times 1) = 2,$ which is to be regarded as not greater than $(2)$, and thus insufficient.
See Binomial Distribution, specifically $\displaystyle \binom{n}{k}p^kq^{(n-k)}.$
Based on the discussion, and the cited article, the desired computation is
$$\left[\binom{4}{4}(0.2)^4 (0.8)^0\right] + \left[\binom{4}{3}(0.2)^3 (0.8)^1\right].$$