Probability of weather on consecutive days.Probability of a cloudy day is .55 Probability of a sunny day is .45A)What is the probability of three consecutive cloudy days, followed by a sunny day?B)What is the probability that exactly 1 out of 4 consecutive days will be sunny?C)What is the probability that at least 1 out of 4 consecutive days will be sunny?I think part A is a Geometric distribution where a success is a sunny day and the number of trials is 3.

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Maggie Bowman · Accepted Answer

Step 1For A, the probability is simply   0.55  ×  0.55  ×  0.55  ×  0.45. For B and C, I would rather you use the binomial distribution, where probability of success (sunny) is 0.45 and number of trials is 4.Step 2Let X be the number of days, out of 4 to be sunny. X follows a binomial distribution with parameters   n  =  4 and   r  =  0.45.

Jorden Pace · Accepted Answer

Step 1The first case is indeed geometric distribution. The general formula to solve any of these kinds of problems however is binomial distribution.The reason why it&#039;s binomial distribution is because not only do you have to figure out the total probability for a particular case (for example: 1 sunny day and 3 rainy days) but you also have to multiply this with the number of ways to construct this case.You should probably know that the number of ways to construct each case is just the binomial coefficient                     (                    n      k                      )            , where n is the total number of days and k is either the number of sunny or rainy days. If not, it&#039;s not hard to just count up the cases for these problems. Just replace that thing with the number of cases you count. Multiplying this with the probabilities, we have the formula                     (                    n      k                      )                  p    k    (  1  −  p      )          (      n      −      k      )      . Finally, we just sum up all of the cases we need.Step 2It&#039;s easy to show that the total probability is 1. The binomial theorem states that the expansion:  (  x  +  y      )    n    =      ∑          k      =      0        n                      (                    n      k                      )                  x    k        y          n      −      k      Setting   x  =  p and   y  =  1  −  p we get:  (  p  +  1  −  p      )    n    =      ∑          k      =      0        n                      (                    n      k                      )                  p    k    (  1  −  p      )          n      −      k        =  1Step 3For part C, we could just add up all the individual cases, but we could also just find the inverse probability and subtract it from 1.  1  −      ∑          k      =      1        n                      (                    n      k                      )                  p    k    (  1  −  p      )          n      −      k        =  1  −      p    0    (  1  −  p      )          n      −      0        =  1  −  (  1  −  p      )    n   And that should pretty much explain all of those problems.

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