An archer shoots repeatedly at a target. On each shot, independently of any other shot, the probability that she hits the bullseye is 0.2. Let X be the number of shots up to and including her first bullseye, so that X Geo(0.2). What is P(X=4), correct to 3 significant figures?

A Level Maths 9709: Geometric Distribution P(X = r) (MCQ)

$X\sim\mathrm{Geo}(0.2)$ , so $p=0.2$ and $q=1-p=0.8$ . “The first bullseye is on the fourth shot” means $r=4$ .

For $X\sim\mathrm{Geo}(p)$ , $P(X=r)=p(1-p)^{r-1}=pq^{r-1}$

Here this means the first three shots must all miss, and the fourth shot must hit: $P(X=4)=0.2\times(0.8)^{3}$

Compute each part: $(0.8)^{3}=0.512$ $P(X=4)=0.2\times0.512=0.1024$

So $P(X=4)\approx\boxed{0.102}$ (to 3 significant figures).

Check: the exact fraction is $0.2\times(0.8)^3=\frac{1}{5}\times\frac{64}{125}=\frac{64}{625}=0.1024$ , which agrees.

B ( $0.0819$ ): this comes from using $(0.8)^{4}$ instead of $(0.8)^{3}$ . The exponent should be $r-1=3$ misses before the hit, not $r=4$ .
C ( $0.0064$ ): this swaps $p$ and $q$ , computing $0.8\times(0.2)^{3}$ instead of $0.2\times(0.8)^{3}$ .
D ( $0.512$ ): this is just $(0.8)^{3}$ with the factor of $p=0.2$ left out entirely.