Respuesta :
Hello!
Here's the simplest way to solve it. (Not the fastest, but what I would consider the most basic)
I managed to get the answer you were given by accepting >2 days in a row as included in that probability.
Let's look at the number of days that it won't rain on to still be accepted as possible to fulfill the requirements.
The minimum has to be 2 days, because if there's only one there's no way it could have rained 2 days in a row. The maximum is 5, because it's possible it didn't rain all 5 days.
Now, let's calculate how many cases work for each scenario of # of days with no rain.
Two days: the two days out of the 5 must be in a row, so therefore there are 4 possible cases. (NNRRR, RNNRR, RRNNR, RRRNN)
Three days: The best way to find this out is to consider the 5 days to be 4, and considering the two days it must be in a row to be stuck together, as just 1 group instead of 2 separate days. (For example, instead of it being NNRRR, I would consider it GRRR, with the G standing for group, because they cannot be separated anyways.) If you were to consider for each position of G, there would be one different position of N, then you could use 3*3 to find the number of cases in this scenario, or 9. (Elaboration: for example, if you had GRRR, there would be 3 places to put the N in place of R, and for each reposition of G there would be 3 more)
Four days: In this case, we can look at it like NNNNR, with 5 different spots to put the R, resulting in 5 different cases.
Five days: This is fairly simple; there is one case, because there's only one arrangement, NNNNN.
Now to calculate the probability.
For each of the separate number of days (2, 3, 4, or 5) there are different probabilities of each of the individual cases happening.
To calculate it, simply take the individual probabilities for each of the scenarios each day (no rain/rain) happening. (if you have no clue what I'm saying, just keep going)
Two days: There would be 2 days of no rain (1/4 chance) and 3 with (3/4 chance) for each of these cases. It would therefore end up being 1/4 * 1/4 * 3/4 * 3/4 * 3/4, which equals 27/1024.
Keep doing the same thing.
Three days: 1/4 * 1/4 * 1/4 * 3/4 * 3/4 = 9/1024
Four days: 1/4 * 1/4 * 1/4 * 1/4 * 3/4 = 3/1024
Five days: 1/4 * 1/4 * 1/4 * 1/4 * 1/4 = 1/1024
Now, there's one more thing to do. The calculations above are for the chances of one case of that number of no rain days happening, but if you look back you'll remember there are multiple cases. Therefore, just multiply the chance by the number of cases.
Two days: 4 possible cases * 27/1024 = 108/1024
Three days: 9 possible cases * 9/1024 = 81/1024
Four days: 5 possible cases * 3/1024 = 15/1024
Five days: 1 case, so just 1/1024
Just add the probabilities up, and then you'll get 205/1024, which when divided, equals roughly 0.2002.
This seems to be the answer pertaining to the situation presented by your teacher, if you were looking at the chance it wouldn't not rain for two days in a row, your solution would be .7998, but in this case, it would be 0.2002. Looks like your teacher may have had some phrasing issues, it'd be a good idea to check in with him.
Hope this helped! If you need some things clarified, feel free to ask, my explanation wasn't the clearest one.
Answer:
The provided answer is incorrect: the actual answer is [tex]\boxed{\frac{205}{1024}} \approx .2002[/tex].
Step-by-step explanation:
We will use recursion and complementary counting. Let [tex]D_n[/tex] be the probability that after [tex]n[/tex] days, it has never failed to rain twice in a row, and on day [tex]n[/tex], it did not rain. Likewise, let [tex]R_n[/tex] be the probability that after [tex]n[/tex] days, it has never failed to rain two days in a row, and on day [tex]n[/tex], it did rain. We will compute [tex]D_5 + R_5[/tex], which is the probability that it never doesn't rain twice in a row. Then, we subtract that from one to find the probability that it does fail to rain twice in a row.
We write formulas for [tex]D_n[/tex] and [tex]R_n[/tex]. If it did not rain yesterday, it must rain today for our condition to hold, while if it did rain yesterday, it can either rain or not rain. Hence, we have:
[tex]D_n = \frac{R_{n-1}}{4}[/tex]
[tex]R_n = \frac{3R_{n-1}+D_{n-1}}{4}[/tex]
Additionally, we clearly have that [tex]D_1 = \frac{1}{4}[/tex] and [tex]R_1 = \frac{3}{4}[/tex]. We compute the values of [tex]D_n[/tex] and [tex]R_n[/tex] below:
[tex]D_2 = \frac{3}{16}, \, R_2 = \frac{3}{4}[/tex]
[tex]D_3 = \frac{3}{16}, \, R_3 = \frac{45}{64}[/tex]
[tex]D_4 = \frac{45}{256}, \, R_4 = \frac{171}{256}[/tex]
[tex]D_5 = \frac{171}{1024}, \, R_5 = \frac{81}{128}[/tex]
Hence, [tex]D_5 + R_5 = \frac{171}{1024} + \frac{81}{128} = \frac{819}{1024}[/tex]. We therefore have that there is a [tex]\frac{819}{1024}[/tex] chance that it never fails to rain twice in a row, so there is a [tex] 1 - \frac{819}{1024} = \boxed{\frac{205}{1024}} \approx 0.2002[/tex] chance that it does fail to rain twice in a row at some point.
