Question 1169076
Let $X$ be the number of attempts you take to pass the driving test, and $Y$ be the number of attempts your friend takes to pass the driving test. We are given that the probability of you passing on any attempt is $p_1 = 0.9$, and the probability of your friend passing on any attempt is $p_2 = 0.8$. The attempts are independent for both individuals.

The number of attempts for each person follows a geometric distribution. The probability mass function for a geometric distribution is given by $P(K=k) = (1-p)^{k-1}p$, where $k$ is the number of trials until the first success, and $p$ is the probability of success on each trial.

In this case, for you:
$P(X=i) = (1-0.9)^{i-1}(0.9) = (0.1)^{i-1}(0.9)$, for $i = 1, 2, 3, \dots$

For your friend:
$P(Y=j) = (1-0.8)^{j-1}(0.8) = (0.2)^{j-1}(0.8)$, for $j = 1, 2, 3, \dots$

We want to find the probability that the total number of attempts is 8, i.e., $P(X+Y=8)$. Since the attempts are independent, we can write this as the sum of the probabilities of all pairs $(i, j)$ such that $i+j=8$:

$P(X+Y=8) = \sum_{i=1}^{7} P(X=i \text{ and } Y=8-i) = \sum_{i=1}^{7} P(X=i) P(Y=8-i)$

Now, we substitute the probability mass functions:
$P(X+Y=8) = \sum_{i=1}^{7} [(0.1)^{i-1}(0.9)] [(0.2)^{(8-i)-1}(0.8)]$
$P(X+Y=8) = \sum_{i=1}^{7} (0.9)(0.8) (0.1)^{i-1} (0.2)^{7-i}$
$P(X+Y=8) = 0.72 \sum_{i=1}^{7} (0.1)^{i-1} (0.2)^{7-i}$

Let $k = i-1$, so as $i$ goes from 1 to 7, $k$ goes from 0 to 6. The sum becomes:
$\sum_{k=0}^{6} (0.1)^{k} (0.2)^{7-k}$
$= (0.1)^0 (0.2)^7 + (0.1)^1 (0.2)^6 + (0.1)^2 (0.2)^5 + (0.1)^3 (0.2)^4 + (0.1)^4 (0.2)^3 + (0.1)^5 (0.2)^2 + (0.1)^6 (0.2)^1$
$= 1 \times 0.0000128 + 0.1 \times 0.000064 + 0.01 \times 0.00032 + 0.001 \times 0.0016 + 0.0001 \times 0.008 + 0.00001 \times 0.04 + 0.000001 \times 0.2$
$= 0.0000128 + 0.0000064 + 0.0000032 + 0.0000016 + 0.0000008 + 0.0000004 + 0.0000002$
$= 0.0000254$

Now, multiply by 0.72:
$P(X+Y=8) = 0.72 \times 0.0000254 = 0.000018288$

Final Answer: The final answer is $\boxed{0.000018288}$