Statistics independence question

Question

A newly-minted sixteen year old takes the written and road drivers's license test repeatedly until he passes them. It is given that the probability he passes the written test is 0.9, that he passes the road test is 0.6 and that the tests are independent. Furthermore, it is assumed that the road test cannot be taken unless he passes the written test, and that once he passes the written test he does not need to take it ever again. Also assume that the written and road tests are distinct attempts. What is the probability that he will pass the road test on his 5th (total) attempt?

I found that since those probability is independent that P(written|road)=P(written)=0.9 then I also know that P(Written intersection road)=P(w)P(R)= (.9)(.6) = .54

I really don't know how to find 5th attempt for this problem..Thank you for any help!

Answer 1

For much larger numbers, we would need to develop general theory. In this case, we don't need to, although we do need to be systematic.

Case 1: Our hero passes the written on the first try, then fails the road test $3$ times, then passes. The probability of this is $(0.9)(0.4)^3(0.6)$.

Case 2: She fails the written, then passes it, then fails the road test twice, then passes it. The probability of this is $(0.1)(0.9)(0.4)^2(0.6)$.

There are a few other cases. The last one has our candidate fail the written three times, then pass, then pass the road test on the first try.

Find the missing probabilities, and add up.

Remark: The calculation assumes not only independence between written and road, but that performances on successive written (or road) tests are independent. In other words, the candidate learns nothing from failure. This is not a reasonable assumption.

Answer 2

Let $W$ denote a pass on the written test, and $R$ denote a pass on the road test. Let $\overline{W},\overline{R}$ denote fails respectively.

The 'allowable' passing sequences are $\overline{W}^{m-1} W \overline{R}^{n-1} R$, and the probability of a particular sequence is $p_{m,n} = (1-0.9)^{m-1} 0.9 (1-0.6)^{n-1} 0.6$.

We want to find $\sum_{m+n=5} p_{m,n}$. (The relevant pairs of $(m,n)$ are $(1,4), (2,3), (3,2), (4,1)$.)

Answer 3

He could pass the written test in attempts $k = 1, 2, 3, 4$. For each, the probability to then pass road test on the fifth attempt is

$$p(k) = (0.1)^{k-1}(0.9)(0.4)^{5-k}(0.6) = \frac{864}{15625 \cdot 4^k}$$

So the probability you seek is
$$\sum_{k = 1}^{4} p(k) = \frac{864}{15625} \sum_{k = 1}^{4} \frac{1}{4^k} \sim 1.84\%$$

The last sum may be easier if you look at it as a G.P.

Answer 4

What is the probability that he would take written test once and take road test exactly four times? $.9\times.4\times.4\times.4\times.6$.

Then what's the probability that he would take written test twice and take road test exactly three times? $.1 \times .9\times.4\times.4\times.6$.

Now continue to take the other possibilities and add the probabilities together.

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Generally, let $W$ be the number of written test taken to pass. Then $W$ follows a geometric distribution and $\Pr(W=w) = .9 \times.1^{w-1}$ for $w\ge1$.

Similarly, let $R$ be the number of road test taken to pass. $\Pr(R=r) = .6 \times.4^{r-1}$ for $r\ge1$.

To get the general probability of $\Pr(W+R=x)$, we find the distribution of $X=W+R$, which is the convolution of the distributions of $W$ and $R$.

$$\Pr(X=x) = \sum_{k=-\infty}^{\infty}\Pr(W=k)\Pr(R=x-k) = \sum_{k=1}^{x-1}\Pr(W=k)\Pr(R=x-k)$$