(i) To find \(P(2 \leq X \leq 6)\), we use the geometric distribution formula. The probability of passing on the \(n\)-th attempt is \((1-p)^{n-1}p\), where \(p = 0.3\).

\(P(2 \leq X \leq 6) = P(X \leq 6) - P(X \leq 1) = 1 - (0.7)^6 - (0.7)\)

Calculating each term:

\((0.7)^6 = 0.117649\)

\((0.7) = 0.7\)

\(P(2 \leq X \leq 6) = 1 - 0.117649 - 0.7 = 0.582351\)

Thus, \(P(2 \leq X \leq 6) \approx 0.582\).

(ii) The expected value \(E(X)\) for a geometric distribution is given by \(\frac{1}{p}\).

\(E(X) = \frac{1}{0.3} = 3.3333\)

Thus, \(E(X) = 3 \frac{1}{3}\).

(a) The probability of obtaining a score of 17 or more in one throw is given by:

\(P(17 \text{ or } 18) = \frac{4}{216} = \frac{1}{54}\)

To find the probability that this score is first obtained on the 6th throw, we use the geometric distribution:

\(P(X = 6) = \left(\frac{53}{54}\right)^5 \times \frac{1}{54}\)

Calculating this gives:

\(P(X = 6) = 0.0169\)

(b) To find the probability that a score of 17 or more is obtained in fewer than 8 throws, we use the complement rule:

\(P(X < 8) = 1 - \left(\frac{53}{54}\right)^7\)

Calculating this gives:

\(P(X < 8) = 0.123\)

An alternative method involves summing probabilities for each throw from 1 to 7:

\(P(X < 8) = \left(\frac{1}{54}\right) + \left(\frac{53}{54}\right)\left(\frac{1}{54}\right) + \left(\frac{53}{54}\right)^2\left(\frac{1}{54}\right) + \ldots + \left(\frac{53}{54}\right)^6\left(\frac{1}{54}\right)\)

This also results in:

\(P(X < 8) = 0.123\)

(a) The probability of obtaining a 4 on a single throw is \(\frac{1}{6}\). The probability of not obtaining a 4 is \(\frac{5}{6}\). For the first 7 throws, Ramesh does not get a 4, and on the 8th throw, he gets a 4. The probability is:

\(\left( \frac{5}{6} \right)^7 \times \frac{1}{6} = \frac{78125}{1679616} \approx 0.0465\)

(b) The probability that it takes no more than 5 throws to obtain a 4 is the complement of not obtaining a 4 in all 5 throws:

\(P(X < 6) = 1 - \left( \frac{5}{6} \right)^5\)

Alternatively, it can be calculated as the sum of probabilities of obtaining a 4 on the 1st, 2nd, 3rd, 4th, or 5th throw:

\(\frac{1}{6} + \left( \frac{5}{6} \right) \left( \frac{1}{6} \right) + \left( \frac{5}{6} \right)^2 \left( \frac{1}{6} \right) + \left( \frac{5}{6} \right)^3 \left( \frac{1}{6} \right) + \left( \frac{5}{6} \right)^4 \left( \frac{1}{6} \right)\)

Calculating gives:

\(\frac{4651}{7776} \approx 0.598\)

(a) The probability that the first lemon chocolate is the 7th is given by the geometric distribution:

\(\left( \frac{4}{5} \right)^6 \times \frac{1}{5} = \frac{4096}{78125} \approx 0.0524\).

(b) The probability that the first lemon chocolate is after at least 6 checks is:

\(1 - \left( \frac{4}{5} \right)^5 = \frac{4096}{15625} \approx 0.262\).

(c) The probability that none of Petra's chocolates is orange is:

\(\frac{10}{15} \times \frac{9}{14} \times \frac{8}{13} = \frac{24}{91} \approx 0.2637\).

(d) The probability that each chocolate has a different flavour is:

\(\frac{7}{15} \times \frac{5}{14} \times \frac{3}{13} = \frac{3}{13} \approx 0.231\).

(e) Given none are orange, the probability that at least 2 are strawberry is:

\(\frac{7}{15} \times \frac{6}{14} \times \frac{5}{13} + \frac{7}{15} \times \frac{6}{14} \times \frac{3}{13} + \frac{7}{15} \times \frac{6}{14} \times \frac{3}{13} = \frac{49}{60} \approx 0.817\).

(b) The probability that the first wet day is 8 October means that the first 7 days are dry and the 8th day is wet. The probability of a dry day is 1 - 0.16 = 0.84. Therefore, the probability is given by:

\((0.84)^7 \times 0.16 = 0.0472\)

(c) We need to find the probability that in exactly 1 out of 4 years, the first wet day is 8 October. Using the result from part (b), the probability for one year is 0.0472. The probability that this happens in exactly 1 out of 4 years is given by the binomial probability formula:

\(4 \times 0.0472 \times (1 - 0.0472)^3 = 0.163\)

(a) The probability of obtaining two tails in one throw is \(\frac{1}{4}\). The probability of not obtaining two tails is \(\frac{3}{4}\). For the first time on the 7th throw, we need 6 failures followed by 1 success:

\(\left( \frac{3}{4} \right)^6 \times \frac{1}{4} = \frac{729}{16384} \approx 0.0445\)

(b) The probability that it takes more than 9 throws is the probability of failing 9 times:

\(\left( \frac{3}{4} \right)^9 = \frac{19683}{262144} \approx 0.0751\)

(a) To find the probability that the score is 4, we need to consider the possible combinations of dice rolls that sum to 4. The possible combinations are (1, 1, 2), (1, 2, 1), and (2, 1, 1). Each die has a probability of \(\frac{1}{6}\) for each face. Therefore, the probability for each combination is \(\left( \frac{1}{6} \right)^3\). Since there are 3 such combinations, the total probability is:

\(3 \times \left( \frac{1}{6} \right)^3 = \frac{3}{216} = \frac{1}{72}\)

(b) The probability of obtaining a score of 18 in a single throw is \(\left( \frac{1}{6} \right)^3 = \frac{1}{216}\) because all dice must show a 6. To find the probability that this occurs for the first time on the 5th throw, we use the geometric distribution formula:

\(P(18 \text{ on 5th throw}) = \left( \frac{215}{216} \right)^4 \times \frac{1}{216}\)

Calculating this gives:

\(\left( \frac{215}{216} \right)^4 \approx 0.981\)

\(0.981 \times \frac{1}{216} \approx 0.00454\)

(a) The mean of a geometric distribution with success probability \(p\) is given by \(\frac{1}{p}\). For a fair die, the probability of rolling a 5 is \(\frac{1}{6}\). Thus, the mean is \(\frac{1}{\frac{1}{6}} = 6\).

(b) The probability that a 5 is first obtained after the 3rd throw but before the 8th throw is calculated as:

\(\left( \frac{5}{6} \right)^3 \cdot \frac{1}{6} + \left( \frac{5}{6} \right)^4 \cdot \frac{1}{6} + \left( \frac{5}{6} \right)^5 \cdot \frac{1}{6} + \left( \frac{5}{6} \right)^6 \cdot \frac{1}{6}\)

Alternatively, it can be calculated as:

\(\left( \frac{5}{6} \right)^3 - \left( \frac{5}{6} \right)^7\)

This results in approximately 0.300.

(c) The probability that a 5 is first obtained in fewer than 10 throws is:

\(1 - \left( \frac{5}{6} \right)^9\)

This results in approximately 0.806.

(a) The probability of not getting a 3 on a single spin is \(\frac{4}{5}\). The probability of getting a 3 on the 8th spin for the first time is \(\left( \frac{4}{5} \right)^7 \times \frac{1}{5}\). This evaluates to \(\frac{16384}{390625}\) or approximately 0.0419.

(b) The probability of getting a 3 for the first time in fewer than 6 spins is the complement of not getting a 3 in the first 5 spins. This is calculated as:

\(1 - \left( \frac{4}{5} \right)^5\)

Alternatively, it can be calculated using the sum of probabilities for getting a 3 on each of the first 5 spins:

\(\frac{1}{5} + \frac{4}{5} \times \frac{1}{5} + \left( \frac{4}{5} \right)^2 \times \frac{1}{5} + \left( \frac{4}{5} \right)^3 \times \frac{1}{5} + \left( \frac{4}{5} \right)^4 \times \frac{1}{5}\)

This evaluates to \(\frac{2101}{3125}\) or approximately 0.672.

(a) The probability of not obtaining a 6 on a single throw is \(\frac{5}{6}\). The probability of not obtaining a 6 for 8 consecutive throws is \(\left( \frac{5}{6} \right)^8\). Therefore, the probability that obtaining a 6 takes more than 8 throws is \(\left( \frac{5}{6} \right)^8 = 0.233\).

(b) For two dice, the probability of obtaining a pair of 6s on a single throw is \(\frac{1}{36}\). The expected number of throws to get a pair of 6s is the reciprocal of this probability, which is \(\frac{1}{\frac{1}{36}} = 36\).

(c) The probability of obtaining a pair of 6s on the 10th throw is \(\left( \frac{35}{36} \right)^9 \times \frac{1}{36}\). The probability of obtaining a pair of 6s on the 11th throw is \(\left( \frac{35}{36} \right)^{10} \times \frac{1}{36}\). Therefore, the probability that obtaining a pair of 6s takes either 10 or 11 throws is \(\left( \frac{35}{36} \right)^9 \times \frac{1}{36} + \left( \frac{35}{36} \right)^{10} \times \frac{1}{36} = 0.0425\).

(a) The probability of getting tails on both coins in one throw is \(\frac{1}{4}\). The expected number of throws to get a pair of tails is the reciprocal of this probability, which is \(\frac{1}{\frac{1}{4}} = 4\).

(b) The probability of getting exactly 3 throws to obtain a pair of tails is calculated by getting not tails in the first two throws and tails in the third throw. The probability is \(\left( \frac{3}{4} \right)^2 \times \frac{1}{4} = \frac{9}{64}\).

(c) The probability that fewer than 6 throws are required is the complement of the probability that 6 or more throws are required. The probability of not getting tails in the first 5 throws is \(\left( \frac{3}{4} \right)^5\). Therefore, the probability of fewer than 6 throws is \(1 - \left( \frac{3}{4} \right)^5 = 0.763\).

The probability of landing on a 5 in one spin is \(P(5) = 0.2\) since the spinner is fair and has 5 sides.

We need to find the probability that it takes fewer than 7 spins to get a 5. This is the complement of not getting a 5 in all 6 spins.

The probability of not getting a 5 in one spin is \(1 - 0.2 = 0.8\).

The probability of not getting a 5 in 6 spins is \(0.8^6\).

Therefore, the probability of getting at least one 5 in fewer than 7 spins is:

\(P(X < 7) = 1 - 0.8^6\)

Calculating this gives:

\(1 - 0.8^6 = 1 - 0.262144 = 0.737856\)

Thus, the probability that it takes fewer than 7 spins for George to obtain a 5 is approximately 0.73856.

(a) The possible combinations to get a score of 4 are (1,3), (3,1), and (2,2). There are 3 favorable outcomes. The total number of outcomes when two dice are thrown is 36. Therefore, the probability is \(\frac{3}{36} = \frac{1}{12}\).

(b) The mean of a geometric distribution is given by \(\frac{1}{p}\), where \(p\) is the probability of success. Here, \(p = \frac{1}{12}\), so the mean is \(\frac{1}{\frac{1}{12}} = 12\).

(c) The probability that the first success occurs on the 6th trial is given by \(\left( \frac{11}{12} \right)^5 \times \frac{1}{12}\). Calculating this gives \(0.0539\) or \(\frac{161051}{2985984}\).

(d) The probability that \(X < 8\) is \(1 - \left( \frac{11}{12} \right)^7\). Calculating this gives \(0.456\) or \(\frac{16344637}{35831808}\).

The probability of obtaining a 1 or a 6 on a single throw of a fair die is \(\frac{2}{6} = \frac{1}{3}\).

The probability of not obtaining a 1 or a 6 on a single throw is \(\frac{4}{6} = \frac{2}{3}\).

We need to find the probability that it takes at least 3 throws but no more than 5 throws to obtain a 1 or a 6.

This can be calculated as:

\(\left( \frac{1}{3} \right) \left( \frac{2}{3} \right)^2 + \left( \frac{1}{3} \right) \left( \frac{2}{3} \right)^3 + \left( \frac{1}{3} \right) \left( \frac{2}{3} \right)^4\)

Calculating each term:

\(\left( \frac{1}{3} \right) \left( \frac{2}{3} \right)^2 = \frac{4}{27}\)

\(\left( \frac{1}{3} \right) \left( \frac{2}{3} \right)^3 = \frac{8}{81}\)

\(\left( \frac{1}{3} \right) \left( \frac{2}{3} \right)^4 = \frac{16}{243}\)

Adding these probabilities:

\(\frac{4}{27} + \frac{8}{81} + \frac{16}{243} = \frac{76}{243}\)

Thus, the probability is \(\frac{76}{243}\) or approximately 0.313.

(a) The sum of probabilities must equal 1: \(0.28 + p + 2p + 3p = 1\). Simplifying gives \(0.28 + 6p = 1\). Solving for \(p\), we get \(6p = 0.72\), so \(p = 0.12\).

(b) For fair spinners (blue and green), the probability of any score is 0.25. We consider scenarios where the sum of scores is 4 or less:

• R = 1, B = 1, G = 1: \(0.28 \times (0.25)^2 = 0.0175\)
• R = 1, B = 1, G = 2: \(0.28 \times (0.25)^2 = 0.0175\)
• R = 1, B = 2, G = 1: \(0.28 \times (0.25)^2 = 0.0175\)
• R = 2, B = 1, G = 1: \(0.12 \times (0.25)^2 = 0.0075\)

Adding these probabilities gives \(0.0175 + 0.0175 + 0.0175 + 0.0075 = 0.06\).

(c) First, find \(P(X \text{ is odd}) = 0.28 + 2p = 0.28 + 0.24 = 0.52\).

Scenarios where the product of scores is 4 or less and X is odd:

• R = 1, B = 1, G = 1: \(0.28 \times (0.25)^2 = 0.0175\)
• R = 1, B = 1, G = 2: \(0.28 \times (0.25)^2 = 0.0175\)
• R = 1, B = 2, G = 1: \(0.28 \times (0.25)^2 = 0.0175\)
• R = 1, B = 1, G = 3: \(0.28 \times (0.25)^2 = 0.0175\)
• R = 1, B = 3, G = 1: \(0.28 \times (0.25)^2 = 0.0175\)
• R = 3, B = 1, G = 1: \(0.24 \times (0.25)^2 = 0.015\)

Total probability for product less than 4 and X odd: \(0.28 \times (0.25)^2 \times 8 + 0.24 \times (0.25)^2 = 0.155\).

Conditional probability: \(\frac{0.155}{0.52} = 0.298\).

To solve these problems, we first need to determine the probability of obtaining a score of 8 or more with two dice. The possible scores are 8, 9, 10, 11, and 12. The combinations for these scores are:

• 8: (2,6), (3,5), (4,4), (5,3), (6,2)
• 9: (3,6), (4,5), (5,4), (6,3)
• 10: (4,6), (5,5), (6,4)
• 11: (5,6), (6,5)
• 12: (6,6)

There are 36 possible outcomes when two dice are thrown. The number of favorable outcomes for a score of 8 or more is 15. Thus, the probability of obtaining a score of 8 or more in one throw is:

\(p = \frac{15}{36} = \frac{5}{12} \approx 0.4167\)

The probability of not obtaining a score of 8 or more is:

\(1 - p = \frac{21}{36} = \frac{7}{12} \approx 0.5833\)

(a) The probability that it takes exactly 5 throws to obtain a score of 8 or more is given by the geometric distribution:

\(P(X = 5) = (1-p)^4 \times p = \left(\frac{21}{36}\right)^4 \times \frac{15}{36}\)

Calculating this gives:

\(P(X = 5) = \frac{12005}{248832} \approx 0.0482\)

(b) The probability that it takes no more than 4 throws to obtain a score of 8 or more is:

Method 1:

\(P(X \leq 4) = 1 - \left(\frac{21}{36}\right)^4\)

Calculating this gives:

\(P(X \leq 4) = \frac{18335}{20736} \approx 0.884\)

Method 2:

Using the cumulative probability:

\(P(X \leq 4) = \frac{15}{36} + \frac{15}{36} \times \frac{21}{36} + \frac{15}{36} \times \left(\frac{21}{36}\right)^2 + \frac{15}{36} \times \left(\frac{21}{36}\right)^3\)

Calculating this gives the same result:

\(P(X \leq 4) = \frac{18335}{20736} \approx 0.884\)

(a) The probability of obtaining a pair of heads (HH) in one throw is \(\frac{1}{4}\). The expected number of throws \(E(X)\) is the reciprocal of this probability, so \(E(X) = \frac{1}{\frac{1}{4}} = 4\).

(b) The probability that exactly 5 throws are required is given by the geometric distribution formula: \(P(X = 5) = \left( \frac{3}{4} \right)^4 \left( \frac{1}{4} \right) = 0.0791\).

(c) The probability that fewer than 7 throws are required is \(P(X < 7) = 1 - \left( \frac{3}{4} \right)^6\). Calculating this gives \(1 - \left( \frac{3}{4} \right)^6 = 0.822\).

(a) To find \(P(X = 4)\), we need the probability that the first 3 spins do not result in a 2, and the 4th spin does. The probability of not getting a 2 on a single spin is \(\frac{4}{5}\), and the probability of getting a 2 is \(\frac{1}{5}\). Therefore,

\(P(X = 4) = \left( \frac{4}{5} \right)^3 \left( \frac{1}{5} \right) = (0.8)^3 (0.2) = 0.1024 = \frac{64}{625}.\)

(b) To find \(P(X < 6)\), we calculate the probability of getting a 2 within the first 5 spins. This is the complement of not getting a 2 in the first 5 spins:

\(P(X < 6) = 1 - (0.8)^5.\)

Calculating this gives:

\(P(X < 6) = 1 - 0.8^5 = 1 - 0.32768 = 0.67232.\)

Let the probability of receiving an elephant be \(\frac{1}{5}\) and the probability of not receiving an elephant be \(\frac{4}{5}\).

We need to find the probability that the first elephant is received before the 6th Monday, i.e., \(P(X < 6)\).

Method 1:

Calculate \(P(X \leq 5) = 1 - (0.8)^5\).

\(P(X \leq 5) = 1 - 0.8^5 = 0.672\).

Method 2:

Calculate \(P(X < 6) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)\).

\(P(X = 1) = \frac{1}{5}\).

\(P(X = 2) = \left( \frac{4}{5} \right) \times \frac{1}{5}\).

\(P(X = 3) = \left( \frac{4}{5} \right)^2 \times \frac{1}{5}\).

\(P(X = 4) = \left( \frac{4}{5} \right)^3 \times \frac{1}{5}\).

\(P(X = 5) = \left( \frac{4}{5} \right)^4 \times \frac{1}{5}\).

Summing these probabilities gives \(0.672\).

(b) The probability that the 5th person is the first person not in favour is given by the geometric distribution formula:

\((0.8)^4 \times (0.2) = 0.08192\)

This is because the first four people are in favour (probability 0.8 each), and the 5th person is not in favour (probability 0.2).

(c) The probability that the 7th person is the second person not in favour is calculated using the negative binomial distribution:

\((0.8)^5 \times (0.2)^2 \times 6 = 0.0786\)

This accounts for 5 people in favour, 2 not in favour, and the combination factor of choosing which two out of seven are not in favour.

(c) The probability of not getting a 3 on a single spin is \(\frac{3}{4}\). The probability of getting a 3 on the 6th spin is:

\(P(Y = 6) = \left( \frac{3}{4} \right)^5 \times \frac{1}{4} = 0.0593\)

(d) The probability of not getting a 3 in the first 4 spins is \(\left( \frac{3}{4} \right)^4\). Therefore, the probability of getting a 3 after more than 4 spins is:

\(P(Y > 4) = \left( \frac{3}{4} \right)^4 = \frac{81}{256} = 0.316\)

Alternatively, \(P(Y > 4) = 1 - P(Y \leq 4)\) where:

\(P(Y \leq 4) = \frac{1}{4} + \frac{3}{4} \times \frac{1}{4} + \left( \frac{3}{4} \right)^2 \times \frac{1}{4} + \left( \frac{3}{4} \right)^3 \times \frac{1}{4}\)

\(P(Y > 4) = 1 - \frac{175}{256} = \frac{81}{256} = 0.316\)