Past Exam Questions

Let \(X\) be the number of students who own a car out of the five chosen. \(X\) follows a binomial distribution with parameters \(n = 5\) and \(p = \frac{1}{4}\).

We need to find \(P(X \geq 4) = P(X = 4) + P(X = 5)\).

Using the binomial probability formula:

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

\(P(X = 5) = \binom{5}{5} \left( \frac{1}{4} \right)^5 \left( \frac{3}{4} \right)^0\)

Calculating each term:

\(P(X = 4) = 5 \times \left( \frac{1}{4} \right)^4 \times \frac{3}{4} = 0.014648\)

\(P(X = 5) = 1 \times \left( \frac{1}{4} \right)^5 = 0.00097656\)

Adding these probabilities gives:

\(P(X \geq 4) = 0.014648 + 0.00097656 = 0.0156\)

Thus, the probability that at least four of these students own a car is \(0.0156\) or \(\frac{1}{64}\).

Let the probability that a person agrees to complete the survey be 0.2. Therefore, the probability that a person does not agree is 0.8.

The probability that none of the first three people agree to complete the survey is given by:

\((0.8)^3 = 0.512\)

Thus, the probability that at least one person agrees is:

\(1 - (0.8)^3 = 1 - 0.512 = 0.488\)

Alternatively, using the binomial probability formula, the probability that exactly one, two, or three people agree can be calculated and summed:

\(P(1, 2, 3) = \binom{3}{1}(0.2)(0.8)^2 + \binom{3}{2}(0.2)^2(0.8) + (0.2)^3\)

\(= 3(0.2)(0.8)^2 + 3(0.2)^2(0.8) + (0.2)^3\)

\(= 0.384 + 0.096 + 0.008 = 0.488\)

(i) The probability that the die lands on 4 is \(\frac{1}{4}\). The probability of getting exactly 2 heads in 4 coin tosses is given by the binomial probability formula:

\(P(4, 2H) = \frac{1}{4} \times \binom{4}{2} \left( \frac{1}{3} \right)^2 \left( \frac{2}{3} \right)^2\)

\(= \frac{1}{4} \times 6 \times \frac{1}{9} \times \frac{4}{9} = \frac{2}{27}\)

(ii) The probability that the die lands on 3 is \(\frac{1}{4}\). The probability of getting exactly 3 heads in 3 coin tosses is:

\(P(3, 3H) = \frac{1}{4} \times \left( \frac{1}{3} \right)^3 = \frac{1}{108}\)

(iii) We need to find the probability that the number the die lands on is the same as the number of heads. This includes the following cases:

• Die lands on 1 and 1 head: \(\frac{1}{4} \times \frac{1}{3} = \frac{1}{12}\)
• Die lands on 2 and 2 heads: \(\frac{1}{4} \times \binom{2}{2} \left( \frac{1}{3} \right)^2 = \frac{1}{36}\)
• Die lands on 3 and 3 heads: \(\frac{1}{4} \times \left( \frac{1}{3} \right)^3 = \frac{1}{108}\)
• Die lands on 4 and 4 heads: \(\frac{1}{4} \times \left( \frac{1}{3} \right)^4 = \frac{1}{324}\)

Summing these probabilities gives:

\(\frac{1}{12} + \frac{1}{36} + \frac{1}{108} + \frac{1}{324} = \frac{10}{81}\)

(i) The two conditions required for a binomial distribution are:

1. Constant probability of success on each trial.
2. Independent trials/events.

(ii) To find the probability that Hebe completes at least 5 puzzles in a week, we use the binomial distribution with parameters:

\(= 7 (number of trials),\)

\(= 0.7 (probability of success).\)

We need to calculate:

\(P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7)\)

\(= \binom{7}{5}(0.7)^5(0.3)^2 + \binom{7}{6}(0.7)^6(0.3)^1 + (0.7)^7\)

\(= 0.647\)

(iii) To find the probability that Hebe completes 4 or fewer puzzles in exactly 3 of the 10 weeks, we first find the probability of completing 4 or fewer puzzles in a week:

\(P(X \leq 4) = 1 - P(X \geq 5) = 1 - 0.647 = 0.3529\)

Now, use the binomial distribution with parameters:

\(= 10 (weeks),\)

\(= 0.3529 (probability of completing 4 or fewer puzzles in a week).\)

We need to calculate:

\(P(Y = 3) = \binom{10}{3}(0.3529)^3(0.6471)^7\)

\(= 0.251\)

Let the probability that Khalid rides his skateboard on a given day be 0.4. We need to find the probability that he rides on his skateboard on at least 2 out of 10 days.

This is a binomial probability problem where the number of trials is 10, and the probability of success (riding the skateboard) is 0.4.

The probability of riding on the skateboard on 0 or 1 day is:

\(P(0) = \binom{10}{0} (0.4)^0 (0.6)^{10} = (0.6)^{10}\)

\(P(1) = \binom{10}{1} (0.4)^1 (0.6)^9 = 10 \times 0.4 \times (0.6)^9\)

Thus, the probability of riding on the skateboard on at least 2 days is:

\(1 - (P(0) + P(1)) = 1 - [(0.6)^{10} + 10 \times 0.4 \times (0.6)^9]\)

Calculating these values:

\((0.6)^{10} \approx 0.0060466176\)

\(10 \times 0.4 \times (0.6)^9 \approx 0.0403936\)

So:

\(1 - (0.0060466176 + 0.0403936) = 1 - 0.04644 \approx 0.954\)

(i) The probability that Annabel eats pasta on exactly 2 days in a week of 7 days can be modeled using the binomial distribution. The probability of eating pasta on a given day is 0.1, and not eating pasta is 0.9. We use the binomial formula:

\(P(2) = \binom{7}{2} (0.1)^2 (0.9)^5\)

Calculating this gives:

\(P(2) = 21 \times 0.01 \times 0.59049 = 0.124\)

(ii) For the second part, we need to find the probability that Annabel eats rice on 2 days, potato on 1 day, and pasta on 2 days in a period of 5 days. The probabilities are 0.75 for rice, 0.15 for potato, and 0.1 for pasta. The number of ways to arrange these days is given by:

\(\frac{5!}{2!1!2!}\)

The probability is then:

\((0.75)^2 (0.15)^1 (0.1)^2 \times \frac{5!}{2!1!2!}\)

Calculating this gives:

\(0.5625 \times 0.15 \times 0.01 \times 30 = 0.0253\)

(i) Let the probability of landing on a 3 be \(p = \frac{1}{3}\). We need to find the probability of getting at least two 3s in four spins. This is given by:

\(P(\geq 2) = 1 - P(0, 1) = 1 - \left(\frac{2}{3}\right)^4 - 4 \cdot \binom{4}{1} \left(\frac{1}{3}\right) \left(\frac{2}{3}\right)^3\)

Alternatively, calculate directly:

\(P(2, 3, 4) = \binom{4}{2} \left(\frac{1}{3}\right)^2 \left(\frac{2}{3}\right)^2 + \binom{4}{3} \left(\frac{1}{3}\right)^3 \left(\frac{2}{3}\right) + \left(\frac{1}{3}\right)^4\)

Thus, \(P(\geq 2) = \frac{11}{27}\) or 0.407.

(ii) To find the probability that the sum of the four scores is 5, consider the combination (1, 1, 1, 2). There are 4 permutations of this combination:

\(P(\text{sum is 5}) = P(1, 1, 1, 2) \times 4 = \left(\frac{1}{3}\right)^4 \times 4\)

Thus, \(P(\text{sum is 5}) = \frac{4}{81}\) or 0.0494.

The probability that Kersley is either asleep or studying at noon on any given day is the sum of the probabilities of these two independent events:

\(P(A) = 0.2 + 0.6 = 0.8\).

We need to find the probability that Kersley is either asleep or studying on at least 6 days out of 7. This is a binomial probability problem with parameters \(n = 7\) and \(p = 0.8\).

The probability of Kersley being asleep or studying on exactly 6 days is given by:

\(P(6) = \binom{7}{6} (0.8)^6 (0.2)^1\).

The probability of Kersley being asleep or studying on all 7 days is:

\(P(7) = (0.8)^7\).

Thus, the probability that Kersley is either asleep or studying on at least 6 days is:

\(P(6 \text{ or } 7) = P(6) + P(7)\).

Calculating these:

\(P(6) = \binom{7}{6} (0.8)^6 (0.2)^1 = 7 \times 0.262144 \times 0.2 = 0.3670016\).

\(P(7) = (0.8)^7 = 0.2097152\).

Therefore,

\(P(6 \text{ or } 7) = 0.3670016 + 0.2097152 = 0.5767168\).

Rounding to three decimal places, the probability is \(0.577\).

(i) We use the binomial probability formula:

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

where \(n = 12\), \(p = 0.65\), and \(1-p = 0.35\).

Calculate for \(r = 8, 9, 10\):

\(\binom{12}{8} (0.65)^8 (0.35)^4 + \binom{12}{9} (0.65)^9 (0.35)^3 + \binom{12}{10} (0.65)^{10} (0.35)^2\)

\(= 0.541\)

(ii) The probability that the fourth passenger is the first to carry a backpack is calculated by considering the first three passengers do not carry a backpack and the fourth does:

\(P(\overline{R}\overline{R}\overline{R}R) = (0.35)^3 \times 0.65\)

\(= 0.0279\)

First, find the probability of throwing a 4. Since the probabilities for scores 5 and 6 are 0.2 each, the total probability for scores 1 to 4 is:

\(1 - 0.4 = 0.6\)

Since scores 1 to 4 have equal probabilities, the probability of throwing a 4 is:

\(\frac{0.6}{4} = 0.15\)

Let \(p = 0.15\) be the probability of throwing a 4, and \(q = 1 - p = 0.85\) be the probability of not throwing a 4.

We need to find the probability of getting a score of 4 on not more than 1 of the 3 throws. This is the sum of the probabilities of getting a score of 4 on exactly 0 or 1 throw:

\(P(0) = q^3 = (0.85)^3\)

\(P(1) = \binom{3}{1} p q^2 = 3 \times 0.15 \times (0.85)^2\)

Thus, the total probability is:

\(P(0) + P(1) = (0.85)^3 + 3 \times 0.15 \times (0.85)^2\)

Calculating these values:

\((0.85)^3 = 0.614125\)

\(3 \times 0.15 \times (0.85)^2 = 0.325875\)

Adding these gives:

\(0.614125 + 0.325875 = 0.939\)

Let the probability that the score is greater than 2 be denoted by \(p\). The score is greater than 2 if the difference between the numbers on the two spinners is 3 or 4. The possible pairs are (1,4), (1,5), (2,5), (4,1), (5,1), (5,2). There are 6 such pairs out of a total of 25 possible outcomes (since each spinner has 5 sides), so \(p = \frac{6}{25}\).

We need to find the probability that the score is greater than 2 on at least 3 occasions out of 9 spins. This is a binomial probability problem where \(X \sim \text{Binomial}(9, \frac{6}{25})\).

The probability that the score is greater than 2 on at least 3 occasions is:

\(P(X \geq 3) = 1 - P(X = 0, 1, 2)\)

Calculate \(P(X = 0, 1, 2)\) using the binomial probability formula:

\(P(X = 0) = \binom{9}{0} \left(\frac{6}{25}\right)^0 \left(\frac{19}{25}\right)^9\)

\(P(X = 1) = \binom{9}{1} \left(\frac{6}{25}\right)^1 \left(\frac{19}{25}\right)^8\)

\(P(X = 2) = \binom{9}{2} \left(\frac{6}{25}\right)^2 \left(\frac{19}{25}\right)^7\)

Substitute and calculate:

\(P(X \geq 3) = 1 - (0.08459 + 0.2404 + 0.3037)\)

\(P(X \geq 3) = 1 - 0.62869 = 0.37131\)

Thus, the probability that the score is greater than 2 on at least 3 occasions is approximately 0.371.

Let the probability that a water pistol does not work properly be denoted by \(p = 0.08\). The probability that a water pistol works properly is \(1 - p = 0.92\).

We are dealing with a binomial distribution where \(n = 19\) and \(p = 0.08\). We need to find the probability that at most 2 do not work properly, i.e., \(P(X \leq 2)\).

The probability \(P(X \leq 2)\) is given by:

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

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

\(P(X = 0) = \binom{19}{0} (0.08)^0 (0.92)^{19} = (0.92)^{19}\)

\(P(X = 1) = \binom{19}{1} (0.08)^1 (0.92)^{18} = 19 \times 0.08 \times (0.92)^{18}\)

\(P(X = 2) = \binom{19}{2} (0.08)^2 (0.92)^{17} = \frac{19 \times 18}{2} \times (0.08)^2 \times (0.92)^{17}\)

Summing these probabilities gives:

\(P(X \leq 2) = (0.92)^{19} + 19 \times 0.08 \times (0.92)^{18} + \frac{19 \times 18}{2} \times (0.08)^2 \times (0.92)^{17}\)

Calculating this expression yields \(P(X \leq 2) = 0.809\).

Let the probability that a car is fitted with satellite navigation equipment be denoted by \(p = 0.76\).

The number of cars fitted with the equipment follows a binomial distribution with parameters \(n = 11\) and \(p = 0.76\).

We need to find \(P(X < 10)\), which is equivalent to \(1 - P(X \geq 10)\).

Calculate \(P(X \geq 10) = P(X = 10) + P(X = 11)\).

Using the binomial probability formula:

\(P(X = 10) = \binom{11}{10} (0.76)^{10} (0.24)^{1}\)

\(P(X = 11) = \binom{11}{11} (0.76)^{11} (0.24)^{0}\)

\(P(X = 10) = 11 \times (0.76)^{10} \times (0.24)\)

\(P(X = 11) = (0.76)^{11}\)

\(P(X \geq 10) = 11 \times (0.76)^{10} \times (0.24) + (0.76)^{11}\)

\(P(X \geq 10) = 0.219\)

Thus, \(P(X < 10) = 1 - 0.219 = 0.781\).

The problem involves a binomial distribution where the number of trials is 10, and the probability of success (rolling a six) is \(\frac{1}{6}\).

We need to find the probability of getting between 3 and 5 sixes inclusive, i.e., \(P(3 \leq X \leq 5)\).

This can be calculated as:

\(P(3, 4, 5) = \binom{10}{3} \left( \frac{1}{6} \right)^3 \left( \frac{5}{6} \right)^7 + \binom{10}{4} \left( \frac{1}{6} \right)^4 \left( \frac{5}{6} \right)^6 + \binom{10}{5} \left( \frac{1}{6} \right)^5 \left( \frac{5}{6} \right)^5\)

Calculating each term:

\(\binom{10}{3} \left( \frac{1}{6} \right)^3 \left( \frac{5}{6} \right)^7 = 0.155045\)

\(\binom{10}{4} \left( \frac{1}{6} \right)^4 \left( \frac{5}{6} \right)^6 = 0.054253\)

\(\binom{10}{5} \left( \frac{1}{6} \right)^5 \left( \frac{5}{6} \right)^5 = 0.012924\)

Adding these probabilities gives:

\(0.155045 + 0.054253 + 0.012924 = 0.222\)

The problem involves a binomial distribution where the probability of success (a household having a printer) is 0.68, and the number of trials is 8. We need to find the probability that 5, 6, or 7 households have a printer.

The probability of exactly k successes in a binomial distribution is given by:

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

where \(n = 8\), \(p = 0.68\), and \(1-p = 0.32\).

Calculate the probabilities for 5, 6, and 7 successes:

\(P(X = 5) = \binom{8}{5} (0.68)^5 (0.32)^3\)

\(P(X = 6) = \binom{8}{6} (0.68)^6 (0.32)^2\)

\(P(X = 7) = \binom{8}{7} (0.68)^7 (0.32)^1\)

Sum these probabilities:

\(P(5, 6, 7) = \binom{8}{5} (0.68)^5 (0.32)^3 + \binom{8}{6} (0.68)^6 (0.32)^2 + \binom{8}{7} (0.68)^7 (0.32)\)

Using the binomial coefficients and calculations:

\(P(5, 6, 7) = 0.722\)

(i) The greatest number of books that could be read is 12. The probability that a member reads this number of books is given by \(P(12) = (0.7)^{12}\).

Calculating this gives \(P(12) = 0.0138\).

(ii) To find the probability that a member reads fewer than 10 books, we calculate \(1 - P(10, 11, 12)\).

Using the binomial probability formula, we have:

\(P(10) = \binom{12}{10} (0.7)^{10} (0.3)^2\)

\(P(11) = \binom{12}{11} (0.7)^{11} (0.3)^1\)

\(P(12) = (0.7)^{12}\)

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

Calculating these probabilities gives \(P(10, 11, 12) = 0.2528\).

Therefore, the probability that a member reads fewer than 10 books is \(1 - 0.2528 = 0.747\).

The problem can be modeled using a binomial distribution where the probability of success (owning a cell phone) is given by \(p = 0.75\), and the number of trials \(n = 8\).

We need to find the probability that the number of successes (adults owning a cell phone) is between 4 and 6 inclusive, i.e., \(P(4 \leq X \leq 6)\).

This can be calculated as:

\(P(4 \leq X \leq 6) = P(X = 4) + P(X = 5) + P(X = 6)\)

Using the binomial probability formula \(P(X = r) = \binom{n}{r} p^r (1-p)^{n-r}\), we calculate each term:

\(P(X = 4) = \binom{8}{4} (0.75)^4 (0.25)^4\)

\(P(X = 5) = \binom{8}{5} (0.75)^5 (0.25)^3\)

\(P(X = 6) = \binom{8}{6} (0.75)^6 (0.25)^2\)

Calculating each:

\(P(X = 4) = 70 \times (0.75)^4 \times (0.25)^4\)

\(P(X = 5) = 56 \times (0.75)^5 \times (0.25)^3\)

\(P(X = 6) = 28 \times (0.75)^6 \times (0.25)^2\)

Summing these probabilities gives:

\(P(4 \leq X \leq 6) = 0.606\)

(i) To find the probability that the numbers on four dice add up to 5, consider the combinations: (1, 1, 1, 2), (1, 1, 2, 1), (1, 2, 1, 1), (2, 1, 1, 1). Each die has a probability of \(\frac{1}{6}\) for a specific number. The probability for one combination is \(\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{1296}\). There are 4 such combinations, so the total probability is \(4 \times \frac{1}{1296} = \frac{1}{324}\).

(ii) Let \(p = \frac{1}{324}\) be the probability from part (i) and \(q = 1 - p = \frac{323}{324}\). We use the binomial probability formula for exactly 1 or 2 successes in 7 trials: \(P(1,2) = \binom{7}{1} p^1 q^6 + \binom{7}{2} p^2 q^5\). Calculate each term: \(\binom{7}{1} \times \frac{1}{324} \times \left(\frac{323}{324}\right)^6 + \binom{7}{2} \times \left(\frac{1}{324}\right)^2 \times \left(\frac{323}{324}\right)^5 = 0.0214\).

Let the random variable \(X\) represent the number of houses with solar heating. \(X\) follows a binomial distribution with parameters \(n = 19\) and \(p = 0.12\), i.e., \(X \sim B(19, 0.12)\).

We need to find \(P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)\).

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

\(P(X = 0) = \binom{19}{0} (0.12)^0 (0.88)^{19} = (0.88)^{19}\)

\(P(X = 1) = \binom{19}{1} (0.12)^1 (0.88)^{18} = 19 \times (0.12) \times (0.88)^{18}\)

\(P(X = 2) = \binom{19}{2} (0.12)^2 (0.88)^{17} = \frac{19 \times 18}{2} \times (0.12)^2 \times (0.88)^{17}\)

\(P(X = 3) = \binom{19}{3} (0.12)^3 (0.88)^{16} = \frac{19 \times 18 \times 17}{6} \times (0.12)^3 \times (0.88)^{16}\)

Summing these probabilities gives:

\(P(X < 4) = (0.88)^{19} + 19 \times (0.12) \times (0.88)^{18} + \frac{19 \times 18}{2} \times (0.12)^2 \times (0.88)^{17} + \frac{19 \times 18 \times 17}{6} \times (0.12)^3 \times (0.88)^{16}\)

Calculating these values, we find \(P(X < 4) = 0.813\).

(i) The three conditions for a binomial distribution are:

1. The probability of success, denoted as \(p\), is constant for each trial.
2. The trials are independent.
3. There are a fixed number of trials, and each trial has only two possible outcomes (success or failure).

(ii) Let \(X\) be the number of months George buys shares in a small company. \(X\) follows a binomial distribution with parameters \(n = 18\) and \(p = 0.15\).

We need to find \(P(X \geq 3)\), which is equivalent to \(1 - P(X \leq 2)\).

Calculate \(P(X = 0)\), \(P(X = 1)\), and \(P(X = 2)\) using the binomial probability formula:

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

\(P(X = 0) = \binom{18}{0} (0.15)^0 (0.85)^{18} = (0.85)^{18}\)

\(P(X = 1) = \binom{18}{1} (0.15)^1 (0.85)^{17} = 18 \times (0.15) \times (0.85)^{17}\)

\(P(X = 2) = \binom{18}{2} (0.15)^2 (0.85)^{16} = \frac{18 \times 17}{2} \times (0.15)^2 \times (0.85)^{16}\)

Thus, \(P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)\)

\(P(X \geq 3) = 1 - P(X \leq 2)\)

Substitute the values:

\(P(X \geq 3) = 1 - [(0.85)^{18} + 18 \times (0.15) \times (0.85)^{17} + \frac{18 \times 17}{2} \times (0.15)^2 \times (0.85)^{16}]\)

\(P(X \geq 3) = 0.520\)