In a certain region, the probability that any given day in October is wet is 0.16, independently of other days.(a) Find the probability that, in a 10-day period in October, fewer than 3 days will be wet.(b) Find the probability that the first wet day in October is 8 October.(c) For 4 randomly chosen years. find the probability that in exactly 1 of these years the first wet day in October is 8 October.

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To solve this problem, we can use the binomial distribution, which gives us the probability of getting a certain number of successes in a fixed number of independent trials.Let X be the number of wet days in a 10-day period in October. Then X follows a binomial distribution with parameters n = 10 (the number of trials) and p = 0.16 (the probability of success).We want to find&amp;nbsp;P(X&amp;lt;3), the probability that fewer than 3 days will be wet. We can calculate this using the cumulative distribution function (CDF) of the binomial distribution:P(X&amp;lt;3)=P(X=0)+P(X=1)+P(X=2)Using the formula for the probability mass function (PMF) of the binomial distribution, we can calculate these probabilities as follows:P(X=k)=(n&amp;nbsp;choose&amp;nbsp;k)⋅pk⋅(1-p)n-kwhere&amp;nbsp;(n&amp;nbsp;choose&amp;nbsp;k)=n!k!⋅(n-k)!&amp;nbsp;is the binomial coefficient.Plugging in the values of n and p, we get:P(X=0)=(10&amp;nbsp;choose&amp;nbsp;0)⋅0.160⋅(1-0.16)10-0=0.326P(X=1)=(10&amp;nbsp;choose&amp;nbsp;1)⋅0.161⋅(1-0.16)10-1=0.384P(X=2)=(10&amp;nbsp;choose&amp;nbsp;2)⋅0.162⋅(1-0.16)10-2=0.230Therefore,P(X&amp;lt;3)=P(X=0)+P(X=1)+P(X=2)=0.326+0.384+0.230=0.940So, the probability that in a 10-day period in October, fewer than 3 days will be wet is 0.940.

In a certain region, the probability that any given

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