the probability that it'll rain 40% over a 3 day period. what is the probability it will rain at least one of the 3 days?

Question

Algebra 2. struggling again.

Answer 1

Hi Austine,

 

The probability of rain over a 3-day period is the product of the probabilities of rain for each day:

    P_3-day = (p_1-day)(P_1-day)(P_1-day)

 

Now let's express this relationship algebraically and then solve for our unknown (x):

    P_1-day = x

    P_3-day = 40% = 40/100 = 0.4 (given)

    (x)(x)(x) = 0.4

    x³ = 0.4

    x = (0.4)^1/3

    x = 0.7368

    x = 0.7368 x 100

    x = 73.68% chance of rain on 1 of the 3 days

 

Thanks for submitting this problem and glad to help.

 

God bless, Jordan. 

Answer 2

To work this problem, let's think about the chances of not raining on each day, which is 100% - 40% = 60%.  So the probability of not raining on any of the three days is .60 * .60 * .60 = .216.  Thus, the chance of raining on at least one of the days is 1 - .216 = .784, or 78.4%.

 

You can also solve this using a tree diagram, which cannot be diagrammed here, but rather the various probabilities are listed:Chance of rain on day 1 only is .4 * .6 * .6 = .144

Day 2 only is the same:  .6 * .4 * .6 = .144

Day 3 only is the same : .6 * .4 * .6 = .144

Day 1 and 2 only is:  .4 * .4 * .6 = .096

Day 1 and 3 are the same:  .4 * .6 * .4 = .096

Day 2 and 3 are the same:  .6 * .4 * .4 = .096

Days 1,2, and 3 are:  .4 * .4 * .4 = .064

Adding these probabilities, you get  .784, or the chance it will rain on at least one of the days.

Again, the chance of not raining at all is .216.  Thus, the sum of the probability of raining at least one the days, and not raining at all, is 1.

Answer 3

If each of the days has a 40% chance of rain, there is a 60% chance of dry weather on each day, or p = 0.6

 

If all days are independent, then the probability of dry weather on all 3 days is p³ = 0.216

 

So it will rain on at least one day with probability

 

1 - 0.216 = 0.784 or 78.4%