# The probability of rain tomorrow is 0.7. The probability of rain the next day is 0.55 and the probability of rain the day after that is 0.4. How do you determine P("it will rain two or more days in the three days")?

Feb 11, 2018

$\frac{577}{1000}$ or $0.577$

#### Explanation:

As probabilities add up to $1$:

First day probability to not rain =$1 - 0.7 = 0.3$

Second day probability to not rain=$1 - 0.55 = 0.45$

Third day probability to not rain=$1 - 0.4 = 0.6$

These are the different possibilities to rain $2$ days:

$R$ means rain, $N R$ means not rain.

color(blue)(P(R, R, NR)) + color(red)(P(R,NR,R)) + color(green)(P(NR, R, R)

Working this out:

color(blue)(P(R,R, NR)=0.7xx0.55xx0.6=231/1000

color(red)(P(R,NR,R)=0.7xx0.45xx0.4=63/500

color(green)(P(NR,R,R)=0.3xx0.55xx0.4=33/500

Probability to rain $2$ days:

$\frac{231}{1000} + \frac{63}{500} + \frac{33}{500}$

Since we need the same denominator we multiply $\frac{63}{500} \mathmr{and} \frac{33}{500}$ by $\frac{2}{2}$:

$\frac{63}{500} \times \frac{2}{2} = \frac{126}{1000}$

$\frac{33}{500} \times \frac{2}{2} = \frac{66}{1000}$

Probability to rain $2$ days:

As denominator is the same, we only add the numerator of the fraction.

$\frac{231}{1000} + \frac{126}{1000} + \frac{66}{1000} = \frac{423}{1000}$

Probability to rain $3$ days:

$P \left(R , R , R\right) = 0.7 \times 0.55 \times 0.4 = \frac{77}{500}$

As the probability to rain over $2$ days is $/ 1000$, we have to change this to $/ 1000$ by $\times \frac{2}{2}$

$\frac{77}{500} \times \frac{2}{2} = \frac{154}{1000}$

Adding all together $P \left(R 2\right) + P \left(R 3\right)$:

$\frac{423}{1000} + \frac{154}{1000} = \frac{577}{1000}$

You could work in decimals if you want, but I find fractions easier to work with. Or you could just convert at the end...

$\frac{577}{1000} = 0.577$

So the probability of rain for $2$ or $3$ days is $\frac{577}{1000}$ or $0.577$

Feb 11, 2018

577/1000 = 0.577 = 57.7%

#### Explanation:

The question is asking for the probability of rain on two or three days. The only situations NOT included are rain on only one day and no rain at all.

Rather than working out all the wanted probabilities, it might be quicker and easier to work out the unwanted probabilities and subtract those from $1$

$P \left(\text{rain on only one day}\right)$

There are 3 options, rain on only the first or the second or third day.

$\textcolor{red}{P \left(R , N , N\right)} + \textcolor{b l u e}{P \left(N , R , N\right)} + \textcolor{g r e e n}{P \left(N , N , R\right)}$

$P \left(\text{no rain") =1-P("rain}\right)$

Fractions are probably easier to use,

$P \left(\text{rain on only one day}\right)$

$= \textcolor{red}{\frac{7}{10} \times \frac{45}{100} \times \frac{6}{10}} + \textcolor{b l u e}{\frac{3}{10} \times \frac{55}{100} \times \frac{6}{10}} + \textcolor{g r e e n}{\frac{3}{10} \times \frac{45}{100} \times \frac{4}{10}}$

$\frac{1890}{10000} + \frac{990}{10000} + \frac{540}{10000} = \frac{3420}{10000}$

$P \left(\text{no rain on any day}\right)$

$= \frac{3}{10} \times \frac{45}{100} \times \frac{6}{10} = \frac{810}{10000}$

$P \left(\text{rain on 2 or 3 days}\right)$

$= \frac{10000}{10000} - \left(\frac{3420}{10000} + \frac{810}{10000}\right) = \frac{5770}{10000}$

$= \frac{577}{1000}$

$= 0.577$

It turns out that one method is not quicker or easier than the other,