Let the probability of passing the test be denoted by \(p = 0.3\) and the probability of failing be \(1 - p = 0.7\). We are looking for the probability that at least three out of five friends pass the test. This can be calculated using the binomial probability formula:

\(P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5)\)

Using the binomial formula \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\), where \(n = 5\):

\(P(X = 3) = \binom{5}{3} (0.3)^3 (0.7)^2 = 10 \times 0.027 \times 0.49 = 0.1323\)

\(P(X = 4) = \binom{5}{4} (0.3)^4 (0.7)^1 = 5 \times 0.0081 \times 0.7 = 0.02835\)

\(P(X = 5) = \binom{5}{5} (0.3)^5 (0.7)^0 = 1 \times 0.00243 \times 1 = 0.00243\)

Adding these probabilities gives:

\(P(X \geq 3) = 0.1323 + 0.02835 + 0.00243 = 0.16308\)

Let the probability of rating the service as 'poor' or 'satisfactory' be the sum of the probabilities of these two ratings:

\(p = 0.12 + 0.36 = 0.48\).

We need to find the probability that more than 2 and fewer than 8 residents rate the service as 'poor' or 'satisfactory'. This is equivalent to finding \(P(3 \leq X \leq 7)\) where \(X\) is a binomial random variable with parameters \(n = 8\) and \(p = 0.48\).

Using the complement rule, we have:

\(P(3 \leq X \leq 7) = 1 - P(X = 0, 1, 2, 8)\)

Calculate \(P(X = 0, 1, 2, 8)\):

\(P(X = 0) = \binom{8}{0} (0.48)^0 (0.52)^8 = 0.00534597\)

\(P(X = 1) = \binom{8}{1} (0.48)^1 (0.52)^7 = 0.039478\)

\(P(X = 2) = \binom{8}{2} (0.48)^2 (0.52)^6 = 0.127544\)

\(P(X = 8) = \binom{8}{8} (0.48)^8 (0.52)^0 = 0.0028179\)

Thus,

\(P(X = 0, 1, 2, 8) = 0.00534597 + 0.039478 + 0.127544 + 0.0028179 = 0.1752\)

Therefore,

\(P(3 \leq X \leq 7) = 1 - 0.1752 = 0.8248\)

Thus, the probability that more than 2 and fewer than 8 residents rate the service as 'poor' or 'satisfactory' is approximately 0.825.

The probability of obtaining a total less than 4 when throwing two dice is given by the outcomes (1,1), (1,2), and (2,1). The probability for each of these outcomes is:

\(\frac{1}{36} + \frac{1}{36} + \frac{1}{36} = \frac{3}{36} = \frac{1}{12}\)

We need to find the probability of obtaining a total less than 4 on at least three throws out of 10. This is a binomial probability problem where:

\(n = 10, \quad p = \frac{1}{12}\)

We need to calculate:

\(1 - P(0, 1, 2)\)

Using the binomial probability formula:

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

Calculate for 0, 1, and 2 successes:

\(P(X = 0) = \binom{10}{0} \left(\frac{1}{12}\right)^0 \left(\frac{11}{12}\right)^{10}\)

\(P(X = 1) = \binom{10}{1} \left(\frac{1}{12}\right)^1 \left(\frac{11}{12}\right)^9\)

\(P(X = 2) = \binom{10}{2} \left(\frac{1}{12}\right)^2 \left(\frac{11}{12}\right)^8\)

Calculate each:

\(P(X = 0) = 0.418904\)

\(P(X = 1) = 0.380822\)

\(P(X = 2) = 0.155791\)

Sum these probabilities:

\(P(0, 1, 2) = 0.418904 + 0.380822 + 0.155791 = 0.955517\)

Therefore, the probability of obtaining a total less than 4 on at least three throws is:

\(1 - 0.955517 = 0.0445\)

Let the probability that a student plays at least one musical instrument be 0.72 (since 28% do not play any instrument).

We need to find the probability that more than 9 students play at least one instrument in a sample of 12 students. This is equivalent to finding:

\(P(X > 9) = P(10) + P(11) + P(12)\)

Using the binomial probability formula:

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

where \(n = 12\), \(p = 0.72\), and \(1-p = 0.28\).

Calculate each probability:

\(P(10) = \binom{12}{10} (0.72)^{10} (0.28)^2\)

\(P(11) = \binom{12}{11} (0.72)^{11} (0.28)^1\)

\(P(12) = \binom{12}{12} (0.72)^{12} (0.28)^0\)

Substitute and calculate:

\(P(10) = 66 \times 0.1934917632 \times 0.0784 = 0.193725\)

\(P(11) = 12 \times 0.1393140695 \times 0.28 = 0.0905726\)

\(P(12) = 1 \times 0.1934917632 \times 1 = 0.0190484\)

Sum these probabilities:

\(P(X > 9) = 0.193725 + 0.0905726 + 0.0190484 = 0.303346\)

Thus, the probability that more than 9 students play at least one musical instrument is approximately 0.304.

(a) To find \(a\), calculate the probability of obtaining exactly 1 head:

\(a = 0.7 \times (0.5)^3 + 0.3 \times (0.5)^3 = \frac{1}{5}\).

For \(b\), calculate the probability of obtaining exactly 2 heads:

\(b = 0.7 \times 0.5^3 \times 3 + 0.3 \times 0.5^3 \times 3 = \frac{3}{8}\).

For \(c\), calculate the probability of obtaining exactly 3 heads:

\(c = 0.7 \times 0.5^3 \times 3 + 0.3 \times 0.5^3 = \frac{3}{10}\).

(b) The expected value \(E(X)\) is calculated as:

\(E(X) = \frac{3 \times 0 + 16 \times 1 + 30 \times 2 + 24 \times 3 + 7 \times 4}{80} = \frac{176}{80} = 2.2\).

(c) The probability of obtaining exactly one head on fewer than 3 occasions is:

\(P(0, 1, 2) = \binom{10}{0} 0.2^0 0.8^{10} + \binom{10}{1} 0.2^1 0.8^9 + \binom{10}{2} 0.2^2 0.8^8\).

Calculating gives \(0.107374 + 0.268435 + 0.301989 = 0.678\).

(d) The probability that Jacob obtains exactly one head for the first time on the 7th or 8th time is:

\(0.8^6 \times 0.2 + 0.8^7 \times 0.2 = 0.0524288 + 0.041943 = 0.09437\).