(a) Let the random variable X be the number of days with more than 10 cm of rain in a 7-day period. X follows a binomial distribution with parameters n = 7 and p = 0.18. We need to find P(X > 2).

\(P(X > 2) = 1 - P(X \leq 2) = 1 - (P(X = 0) + P(X = 1) + P(X = 2)).\)

Using the binomial probability formula, we calculate:

\(P(X = 0) = \binom{7}{0} (0.18)^0 (0.82)^7 = 0.249285,\)

\(P(X = 1) = \binom{7}{1} (0.18)^1 (0.82)^6 = 0.383048,\)

\(P(X = 2) = \binom{7}{2} (0.18)^2 (0.82)^5 = 0.252251.\)

\(Thus, P(X \leq 2) = 0.249285 + 0.383048 + 0.252251 = 0.884584.\)

\(Therefore, P(X > 2) = 1 - 0.884584 = 0.115.\)

(b) Let Y be the number of 7-day periods with at least one day of more than 10 cm of rain. The probability of at least one day of rain in a 7-day period is 1 - P(X = 0) = 1 - 0.82^7 = 0.7507.

\(Y follows a binomial distribution with parameters n = 3 and p = 0.7507. We need to find P(Y = 2).\)

\(P(Y = 2) = \binom{3}{2} (0.7507)^2 (1 - 0.7507)^1 = 3 \times 0.7507^2 \times 0.2493 = 0.421.\)

(a) The sum of probabilities must equal 1: \(p + q + 0.65 = 1\). Therefore, \(p + q = 0.35\).

Using the expected value: \(E(X) = 0 \times 0.6 + 1 \times p + 2 \times q + 3 \times 0.05 = 0.55\).

This simplifies to \(p + 2q + 0.15 = 0.55\), so \(p + 2q = 0.4\).

Solving the equations \(p + q = 0.35\) and \(p + 2q = 0.4\), we find \(q = 0.05\) and \(p = 0.3\).

(c) The probability that \(X = 1\) in at least 3 of 12 turns is \(1 - P(0, 1, 2)\).

\(P(0, 1, 2) = \binom{12}{0} (0.3)^0 (0.7)^{12} + \binom{12}{1} (0.3)^1 (0.7)^{11} + \binom{12}{2} (0.3)^2 (0.7)^{10}\).

Calculating these gives \(0.01384 + 0.07118 + 0.16779 = 0.25299\).

Thus, the probability is \(1 - 0.25299 = 0.747\).

(d) The probability that Jim first succeeds on the 9th turn is \((0.95)^8 \times 0.05\).

This evaluates to \(0.0332\).

Let the random variable \(X\) represent the number of wet days in a 10-day period. \(X\) follows a binomial distribution with parameters \(n = 10\) and \(p = 0.16\).

We need to find \(P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)\).

Using the binomial probability formula:

\(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)

Calculate each probability:

\(P(X = 0) = \binom{10}{0} (0.16)^0 (0.84)^{10} = 0.17490\)

\(P(X = 1) = \binom{10}{1} (0.16)^1 (0.84)^9 = 0.333145\)

\(P(X = 2) = \binom{10}{2} (0.16)^2 (0.84)^8 = 0.28555\)

Summing these probabilities gives:

\(P(X < 3) = 0.17490 + 0.333145 + 0.28555 = 0.794\)

Let the probability that a household has no broadband service be 0.35. We are choosing 10 households, so this is a binomial distribution with parameters \(n = 10\) and \(p = 0.35\).

We need to find the probability that fewer than 3 households have no broadband service, i.e., \(P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)\).

Using the binomial probability formula:

\(P(X = r) = \binom{n}{r} p^r (1-p)^{n-r}\)

Calculate each probability:

\(P(X = 0) = \binom{10}{0} (0.35)^0 (0.65)^{10} = 1 \times 1 \times 0.013463 = 0.013463\)

\(P(X = 1) = \binom{10}{1} (0.35)^1 (0.65)^9 = 10 \times 0.35 \times 0.072492 = 0.072492\)

\(P(X = 2) = \binom{10}{2} (0.35)^2 (0.65)^8 = 45 \times 0.1225 \times 0.178565 = 0.17565\)

Add these probabilities:

\(P(X < 3) = 0.013463 + 0.072492 + 0.17565 = 0.261605\)

Rounding to three decimal places, the probability is \(0.262\).

(a) The probability that the flight does not arrive late is the sum of the probabilities that it arrives early or on time:

\(P( ext{not late}) = 0.15 + 0.55 = 0.7\)

The probability that on each of 3 days the flight does not arrive late is:

\((0.7)^3 = 0.343\)

(b) We need to find the probability that the flight arrives early at least 3 times out of 9 days. This is a binomial probability problem where \(n = 9\), \(p = 0.15\), and we want \(P(X \geq 3)\).

First, calculate \(P(X < 3)\):

\(P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)\)

Using the binomial probability formula \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\), we find:

\(P(X = 0) = \binom{9}{0} (0.15)^0 (0.85)^9 = 0.231617\)

\(P(X = 1) = \binom{9}{1} (0.15)^1 (0.85)^8 = 0.367862\)

\(P(X = 2) = \binom{9}{2} (0.15)^2 (0.85)^7 = 0.259667\)

Thus,

\(P(X < 3) = 0.231617 + 0.367862 + 0.259667 = 0.859146\)

Therefore,

\(P(X \geq 3) = 1 - P(X < 3) = 1 - 0.859146 = 0.140854\)

Rounding to three decimal places, the probability is 0.141.