A certain city has one chance in two of receiving rain on June 1, one chance in five of receiving rain on July 1, and two chances in three of receiving rain on August 1. What is the probability that the city will receive rain on none of these days? (A) 2/15(B) 2/5(C)14/15

Answer:So the probability that the city won't receive rain on none of these days is [tex]\frac{2}{15}[/tex]Step-by-step explanation:From the question we know that the probability of rain on June 1 is [tex]\frac{1}{2}[/tex], on July 1 is [tex]\frac{1}{5}[/tex] and in August 1 is [tex]\frac{2}{3}[/tex], So the probability of no rain at any of these days is calculate as:For June 1:[tex]P(no-rain)=1-\frac{1}{2} =\frac{1}{2}[/tex]For July 1:[tex]P(no-rain)=1-\frac{1}{5} =\frac{4}{5}[/tex]For August 1:[tex]P(no-rain)=1-\frac{2}{3} =\frac{1}{3}[/tex]  So the probability that the city won't receive rain on none of these days is the multiplication of the probabilities of no rain for every one of these days. Then:[tex]P=\frac{1}{2} *\frac{4}{5} *\frac{1}{3} =\frac{4}{30}[/tex]Simplified the value of P, we obtain: [tex]P=\frac{2}{15}[/tex]