Respuesta :
Answer:
[tex]\frac{7}{27}[/tex]
Step-by-step explanation:
Let the event of raining on a day be success.
Then probability of success, [tex]p=\frac{2}{3}[/tex]
So, probability of failure is not raining on a day and is given as,
[tex]q=1-p=1-\frac{2}{3}=\frac{3-2}{3}=\frac{1}{3}[/tex]
Now, number of days to check the probability of rain, [tex]n=3[/tex]
Now, probability of raining at most one day means number of success is either 0 or 1.
Using Bernoulli's distribution for [tex]X=0,1[/tex]
[tex]P(X=0\textrm{ or }X=1)=_{0}^{3}\textrm{C}p^{0}q^{3-0}+_{1}^{3}\textrm{C}p^{1}q^{3-1}\\\\P(X=0\textrm{ or }X=1)=(\frac{1}{3})^{3}+3(\frac{2}{3})^{1}(\frac{1}{3})^{2}\\\\P(X=0\textrm{ or }X=1)=\frac{1}{27}+\frac{2}{9}\\\\P(X=0\textrm{ or }X=1)=\frac{1}{27}+\frac{6}{27}\\\\P(X=0\textrm{ or }X=1)=\frac{1+6}{27}=\frac{7}{27}[/tex]
Therefore, the probability of at most one day of rain during the next three days is [tex]\frac{7}{27}[/tex]
