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)?

2 Answers
Feb 11, 2018

5771000 or 0.577

Explanation:

As probabilities add up to 1:

First day probability to not rain =10.7=0.3

Second day probability to not rain=10.55=0.45

Third day probability to not rain=10.4=0.6

These are the different possibilities to rain 2 days:

R means rain, NR means not rain.

P(R,R,NR)+P(R,NR,R)+P(NR,R,R)

Working this out:

P(R,R,NR)=0.7×0.55×0.6=2311000

P(R,NR,R)=0.7×0.45×0.4=63500

P(NR,R,R)=0.3×0.55×0.4=33500

Probability to rain 2 days:

2311000+63500+33500

Since we need the same denominator we multiply 63500and33500 by 22:

63500×22=1261000

33500×22=661000

Probability to rain 2 days:

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

2311000+1261000+661000=4231000

Probability to rain 3 days:

P(R,R,R)=0.7×0.55×0.4=77500

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

77500×22=1541000

Adding all together P(R2)+P(R3):

4231000+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...

5771000=0.577

So the probability of rain for 2 or 3 days is 5771000 or 0.577

Feb 11, 2018

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 1

P(rain on only one day)

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

P(R,N,N)+P(N,R,N)+P(N,N,R)

P(no rain)=1P(rain)

Fractions are probably easier to use,

P(rain on only one day)

=710×45100×610+310×55100×610+310×45100×410

189010000+99010000+54010000=342010000

P(no rain on any day)

=310×45100×610=81010000

P(rain on 2 or 3 days)

=1000010000(342010000+81010000)=577010000

=5771000

=0.577

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