Question 1169076: You and your friend are taking the driving test to obtain your driver’s licenses. The
probability for you to pass the test is 0.9 and for your friend it is 0.8. Calculate the
probability that you two make a total of 8 attempts to obtain your licenses.
(if you got your license on the 2nd attempt and your friend got his on 3rd attempt, you
made a total of 5 attempts)
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! 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 $\boxed{0.000018288}$
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