Past Exam Questions

(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:

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

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

Let the random variable \(X\) represent the weight of a large bag of pasta. \(X\) is normally distributed with mean \(\mu = 1.5\) kg and standard deviation \(\sigma = 0.05\) kg.

We need to find \(P(1.42 < X < 1.52)\).

Standardize the variable using the formula:

\(Z = \frac{X - \mu}{\sigma}\)

For \(X = 1.42\):

\(Z = \frac{1.42 - 1.5}{0.05} = -1.6\)

For \(X = 1.52\):

\(Z = \frac{1.52 - 1.5}{0.05} = 0.4\)

Thus, \(P(1.42 < X < 1.52) = P(-1.6 < Z < 0.4)\).

Using standard normal distribution tables or a calculator, find:

\(\Phi(0.4) = 0.6554\)

\(\Phi(-1.6) = 0.0548\)

Therefore, \(P(-1.6 < Z < 0.4) = \Phi(0.4) - \Phi(-1.6) = 0.6554 - 0.0548 = 0.6006\).

Rounding to three decimal places, the probability is \(0.601\).

(a) To find the probability that Raj runs for more than 43.2 minutes, we use the standard normal distribution. First, we calculate the z-score:

\(z = \frac{43.2 - 41.2}{3.6} = 0.5556\)

We need to find \(P(Z > 0.5556)\). Using the standard normal distribution table, \(\Phi(0.5556) = 0.7108\).

Therefore, \(P(Z > 0.5556) = 1 - 0.7108 = 0.2892\).

So, the probability that Raj runs for more than 43.2 minutes is approximately 0.289.

(b) To find the number of days Raj runs for less than 43.2 minutes, we calculate:

\(P(X < 43.2) = 1 - P(X > 43.2) = 1 - 0.2892 = 0.7108\).

Multiply this probability by the number of days in a year:

\(0.7108 \times 365 = 259.4\).

Therefore, the estimated number of days is approximately 259 or 260.

(a) To find the probability that a randomly chosen employee takes more than 28.6 minutes, we use the standard normal distribution. First, we calculate the z-score:

\(z = \frac{28.6 - 32.2}{9.6} = -0.375\)

We need \(P(Z > -0.375)\). Using the standard normal distribution table, \(\Phi(-0.375) = 0.3538\). Therefore, \(P(Z > -0.375) = 1 - 0.3538 = 0.6462\).

Thus, the probability is approximately 0.646.

(c) We need to find the probability that the time differs from the mean by less than 15.0 minutes, i.e., \(|X - 32.2| < 15\).

This is equivalent to \(17.2 < X < 47.2\). We calculate the z-scores:

\(z_1 = \frac{17.2 - 32.2}{9.6} = -1.5625\)

\(z_2 = \frac{47.2 - 32.2}{9.6} = 1.5625\)

We need \(P(-1.5625 < Z < 1.5625)\). Using the standard normal distribution table, \(\Phi(1.5625) = 0.9409\).

Thus, \(P(-1.5625 < Z < 1.5625) = 2 \times 0.9409 - 1 = 0.8818\).

Therefore, the probability is approximately 0.882.

(i) To find the number of days Karli spends more than 142 minutes on social media, we first calculate the probability that she spends more than 142 minutes on a given day. We use the standard normal distribution:

\(P(X > 142) = P\left( Z > \frac{142 - 125}{24} \right) = P(Z > 0.7083)\)

Using the standard normal distribution table, \(P(Z > 0.7083) = 1 - 0.7604 = 0.2396\).

Therefore, the expected number of days is \(0.2396 \times 365 = 87.454\).

Rounding to the nearest whole number, we get 87 or 88 days.

(ii) We need to find the probability that Karli spends more than 142 minutes on social media on fewer than 2 out of 10 days. This is a binomial probability problem with \(n = 10\) and \(p = 0.2396\).

\(P(0, 1) = 0.7604^{10} + \binom{10}{1} \times 0.2396^1 \times 0.7604^9\)

\(= 0.064628 + 0.20364\)

\(= 0.268\)

We need to find the probability that a rod's length is within 0.5 cm of the mean, i.e., between 24.7 cm and 25.7 cm.

Using the standardization formula, we calculate:

\(P\left( \frac{25.2 - (25.2 + 0.5)}{0.4} < z < \frac{25.2 - (25.2 - 0.5)}{0.4} \right)\)

This simplifies to:

\(P\left( \frac{-0.5}{0.4} < z < \frac{0.5}{0.4} \right) = P(-1.25 < z < 1.25)\)

Using the standard normal distribution table, we find:

\(P(-1.25 < z < 1.25) = 2 \Phi(1.25) - 1\)

\(\Phi(1.25) \approx 0.8944\)

Thus, \(2 \times 0.8944 - 1 = 0.7888\)

The expected number of rods is:

\(0.7888 \times 500 = 394.4\)

Rounding gives 394 or 395 rods.

Let the random variable representing the time spent be denoted by \(X\), where \(X \sim N(96, 18^2)\).

We need to find \(P(85 < X < 100)\).

Standardize the variable: \(Z = \frac{X - 96}{18}\).

Thus, \(P(85 < X < 100) = P\left(\frac{85 - 96}{18} < Z < \frac{100 - 96}{18}\right)\).

Calculate the standardized values: \(\frac{85 - 96}{18} = -0.6111\) and \(\frac{100 - 96}{18} = 0.2222\).

Therefore, \(P(-0.6111 < Z < 0.2222) = \Phi(0.2222) + \Phi(0.6111) - 1\).

Using standard normal distribution tables, \(\Phi(0.2222) = 0.5879\) and \(\Phi(0.6111) = 0.7294\).

Thus, \(P(-0.6111 < Z < 0.2222) = 0.5879 + 0.7294 - 1 = 0.3173\).

Rounding to three decimal places, the probability is \(0.317\).

Let the random variable \(X\) represent the time taken to swim 100 metres. \(X\) follows a normal distribution with mean \(\mu = 62\) and standard deviation \(\sigma = 5\).

We need to find \(P(56 < X < 66)\).

Standardize the variable using the formula:

\(Z = \frac{X - \mu}{\sigma}\)

For \(X = 56\):

\(Z = \frac{56 - 62}{5} = -1.2\)

For \(X = 66\):

\(Z = \frac{66 - 62}{5} = 0.8\)

Thus, \(P(56 < X < 66) = P(-1.2 < Z < 0.8)\).

Using the standard normal distribution table, find:

\(\Phi(0.8) = 0.7881\)

\(\Phi(1.2) = 0.8849\)

Therefore, \(P(-1.2 < Z < 0.8) = \Phi(0.8) + \Phi(1.2) - 1 = 0.7881 + 0.8849 - 1 = 0.673\).