The probability of rain tomorrow is 0.70.7. The probability of rain the next day is 0.550.55 and the probability of rain the day after that is 0.40.4. How do you determine P("it will rain two or more days in the three days")P(it will rain two or more days in the three days)?

2 Answers
Feb 11, 2018

577/10005771000 or 0.5770.577

Explanation:

As probabilities add up to 11:

First day probability to not rain =1-0.7=0.310.7=0.3

Second day probability to not rain=1-0.55=0.4510.55=0.45

Third day probability to not rain=1-0.4=0.610.4=0.6

These are the different possibilities to rain 22 days:

RR means rain, NRNR means not rain.

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

Working this out:

color(blue)(P(R,R, NR)=0.7xx0.55xx0.6=231/1000P(R,R,NR)=0.7×0.55×0.6=2311000

color(red)(P(R,NR,R)=0.7xx0.45xx0.4=63/500P(R,NR,R)=0.7×0.45×0.4=63500

color(green)(P(NR,R,R)=0.3xx0.55xx0.4=33/500P(NR,R,R)=0.3×0.55×0.4=33500

Probability to rain 22 days:

231/1000+63/500+33/5002311000+63500+33500

Since we need the same denominator we multiply 63/500 and 33/50063500and33500 by 2/222:

63/500xx2/2=126/100063500×22=1261000

33/500xx2/2=66/100033500×22=661000

Probability to rain 22 days:

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

231/1000+126/1000+66/1000=423/10002311000+1261000+661000=4231000

Probability to rain 33 days:

P(R,R,R)=0.7xx0.55xx0.4=77/500P(R,R,R)=0.7×0.55×0.4=77500

As the probability to rain over 22 days is /1000/1000, we have to change this to /1000/1000 by xx2/2×22

77/500xx2/2=154/100077500×22=1541000

Adding all together P(R 2 )+P(R 3)P(R2)+P(R3):

423/1000+154/1000=577/10004231000+1541000=5771000

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

577/1000=0.5775771000=0.577

So the probability of rain for 22 or 33 days is 577/10005771000 or 0.5770.577

Feb 11, 2018

577/1000 = 0.577 = 57.7%5771000=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 11

P("rain on only one day")P(rain on only one day)

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

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

P("no rain") =1-P("rain")P(no rain)=1P(rain)

Fractions are probably easier to use,

P("rain on only one day")P(rain on only one day)

=color(red)(7/10 xx45/100 xx 6/10)+ color(blue)(3/10xx55/100xx6/10)+color(green)(3/10xx45/100xx4/10)=710×45100×610+310×55100×610+310×45100×410

1890/10000 +990/10000+540/10000 = 3420/10000189010000+99010000+54010000=342010000

P("no rain on any day")P(no rain on any day)

= 3/10xx45/100xx6/10 = 810/10000=310×45100×610=81010000

P("rain on 2 or 3 days")P(rain on 2 or 3 days)

= 10000/10000-(3420/10000 +810/10000)= 5770/10000=1000010000(342010000+81010000)=577010000

=577/1000=5771000

=0.577=0.577

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