Question

If the probability that it doesn't rain on a given day is .25, what is the probability that it doesn't rain 2 days in a row during a 5-day period?

The answer is 0.7998! I need to know how to arrive at this answer.​

Answer 1

$\red{\bold{\underline{{ Answer \: with\: Explanation \: :}}}}$

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The Provided answer is incorrect: the actual answer is $\boxed{\frac{205}{1024}} \approx .2002$.

We will use recursion and complementary counting. Let $D_n$ be the probability that after $n$ days, it has never failed to rain twice in a row, and on day $n$, it did not rain.  Likewise, let $R_n$ be the probability that after $n$ days, it has never failed to rain two days in a row, and on day $n$, it did rain.  We will compute $D_5 + R_5$, 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 $D_n$ and $R_n$.  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:

$D_n = \frac{R_{n-1}}{4}$

$R_n = \frac{3R_{n-1}+D_{n-1}}{4}$

Additionally, we clearly have that $D_1 = \frac{1}{4}$ and $R_1 = \frac{3}{4}$.  We compute the values of $D_n$ and $R_n$ below:

$D_2 = \frac{3}{16}, \, R_2 = \frac{3}{4}$

$D_3 = \frac{3}{16}, \, R_3 = \frac{45}{64}$

$D_4 = \frac{45}{256}, \, R_4 = \frac{171}{256}$

$D_5 = \frac{171}{1024}, \, R_5 = \frac{81}{128}$

Hence, $D_5 + R_5 = \frac{171}{1024} + \frac{81}{128} = \frac{819}{1024}$.  We therefore have that there is a $\frac{819}{1024}$ chance that it never fails to rain twice in a row, so there is a $1 - \frac{819}{1024} = \boxed{\frac{205}{1024}} \approx 0.2002$ chance that it does fail to rain twice in a row at some point.

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