Past Exam Questions

The probability that a mango is medium or large is given by:

\(p = 0.70 + 0.15 = 0.85\)

We need to find the probability that fewer than 12 mangoes are medium or large. This is equivalent to finding:

\(P(X < 12) = 1 - P(X \geq 12)\)

Where \(X\) is the number of medium or large mangoes out of 14. We use the binomial distribution:

\(P(X \geq 12) = P(X = 12) + P(X = 13) + P(X = 14)\)

Using the binomial probability formula:

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

Calculate each probability:

\(P(X = 12) = \binom{14}{12} (0.85)^{12} (0.15)^2\)

\(P(X = 13) = \binom{14}{13} (0.85)^{13} (0.15)^1\)

\(P(X = 14) = \binom{14}{14} (0.85)^{14} (0.15)^0\)

Summing these gives:

\(P(X \geq 12) = 0.6479\)

Thus,

\(P(X < 12) = 1 - 0.6479 = 0.352\)

(b) The probability of getting a leopard in one magazine is \(p = 0.2\). We need to find the probability of getting more than two leopards in 12 magazines. This is given by:

\(1 - P(0, 1, 2) = 1 - \left( \binom{12}{0} (0.2)^0 (0.8)^{12} + \binom{12}{1} (0.2)^1 (0.8)^{11} + \binom{12}{2} (0.2)^2 (0.8)^{10} \right)\)

\(= 1 - (0.06872 + 0.20615 + 0.28347)\)

\(= 0.442\)

(c) The probability that after 5 weeks Sahim has exactly one of each animal is given by:

\((0.2)^5 \times 5!\)

\(= 0.0384 = \frac{24}{625}\)

The problem involves a binomial distribution where the probability of being an office worker is given as 0.22. The probability of not being an office worker is therefore 1 - 0.22 = 0.78.

We need to find the probability that between 4 and 6 people inclusive are office workers out of 8 people. This is calculated using the binomial probability formula:

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

where \(n = 8\), \(p = 0.22\), and \(r\) is the number of office workers.

Calculate for \(r = 4, 5, 6\):

\(P(4) = \binom{8}{4} (0.22)^4 (0.78)^4\)

\(P(5) = \binom{8}{5} (0.22)^5 (0.78)^3\)

\(P(6) = \binom{8}{6} (0.22)^6 (0.78)^2\)

Sum these probabilities:

\(P(4, 5, 6) = P(4) + P(5) + P(6)\)

\(= \binom{8}{4} (0.22)^4 (0.78)^4 + \binom{8}{5} (0.22)^5 (0.78)^3 + \binom{8}{6} (0.22)^6 (0.78)^2\)

\(= 0.0763\)

(i) The number of fireworks that fail follows a binomial distribution with parameters \(n = 20\) and \(p = 0.05\). We need to find \(P(X > 1)\).

\(P(X > 1) = 1 - P(X = 0) - P(X = 1)\)

\(P(X = 0) = (0.95)^{20}\)

\(P(X = 1) = \binom{20}{1} (0.95)^{19} (0.05)^1\)

\(P(X > 1) = 1 - (0.95)^{20} - \binom{20}{1} (0.95)^{19} (0.05)\)

\(P(X > 1) = 1 - 0.3585 - 0.3770 = 0.264\)

(ii) The profit if 19 or 20 fireworks work is \(450 \times 10 - 480 = 4020\).

If more than 1 firework fails, the profit is \(-480\).

The expected profit is:

\(4020 \times (1 - 0.264) - 480 \times 0.264\)

\(= 4020 \times 0.736 - 480 \times 0.264\)

\(= 2960.64 - 126.72 = 2833.92\)

Rounding gives \(\$2830\).

The total number of pairs of shoes is 20, and the number of pairs with designer labels is 8. Therefore, the probability of choosing a pair with designer labels is \(p = \frac{8}{20} = 0.4\).

The probability of choosing a pair without designer labels is \(1 - p = 0.6\).

We need to find the probability that Suzanne wears a pair of shoes with designer labels on at most 4 days out of 7. This is a binomial probability problem where \(n = 7\), \(p = 0.4\), and we want \(P(X \leq 4)\).

Using the complement rule, \(P(X \leq 4) = 1 - P(X = 5) - P(X = 6) - P(X = 7)\).

Calculate each probability:

\(P(X = 5) = \binom{7}{5} (0.4)^5 (0.6)^2 = 21 \times 0.01024 \times 0.36 = 0.0774144\).

\(P(X = 6) = \binom{7}{6} (0.4)^6 (0.6)^1 = 7 \times 0.004096 \times 0.6 = 0.0172032\).

\(P(X = 7) = \binom{7}{7} (0.4)^7 (0.6)^0 = 1 \times 0.0016384 \times 1 = 0.0016384\).

Thus, \(P(X \leq 4) = 1 - (0.0774144 + 0.0172032 + 0.0016384) = 1 - 0.096256 = 0.903744\).

Rounding to three decimal places, the probability is \(0.904\).

(i) Let the random variable \(X\) represent the number of days with more than 20 cm of snow in a 7-day period. \(X\) follows a binomial distribution with parameters \(n = 7\) and \(p = 0.21\).

We need to find \(P(X < 5)\), which is \(1 - P(X \geq 5)\).

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

\(P(X = 5) = \binom{7}{5} (0.21)^5 (0.79)^2\)

\(P(X = 6) = \binom{7}{6} (0.21)^6 (0.79)^1\)

\(P(X = 7) = \binom{7}{7} (0.21)^7\)

\(P(X \geq 5) = 21(0.21)^5(0.79)^2 + 7(0.21)^6(0.79) + (0.21)^7\)

\(P(X \geq 5) = 0.006 + 0.0003 + 0.0000008 \approx 0.0063\)

Thus, \(P(X < 5) = 1 - 0.0063 = 0.994\).

(ii) Let \(Y\) be the number of 7-day periods with at least 1 day of more than 20 cm of snow. The probability of at least 1 day with more than 20 cm of snow in a 7-day period is \(1 - (0.79)^7 = 0.808\).

\(Y\) follows a binomial distribution with parameters \(n = 4\) and \(p = 0.808\).

We need to find \(P(Y = 3)\).

\(P(Y = 3) = \binom{4}{3} (0.808)^3 (0.192)^1\)

\(P(Y = 3) = 4(0.808)^3(0.192)\)

\(P(Y = 3) = 4(0.527)(0.192) \approx 0.405\).

First, calculate the probability of a person being group O+ given they are Rhesus +. The probability of being O+ is 37%, and the probability of being Rhesus + is the sum of A+, B+, AB+, and O+, which is 35% + 8% + 3% + 37% = 83%.

Thus, the probability of being O+ given Rhesus + is:

\(P(O \text{ given } +) = \frac{0.37}{0.83} = 0.4458\)

Now, use the binomial distribution to find the probability that fewer than 3 out of 9 people are group O+.

Let \(X\) be the number of people who are group O+. Then \(X \sim \text{Binomial}(9, 0.4458)\).

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

\(P(X = 0) = \binom{9}{0} (0.4458)^0 (0.5542)^9\)

\(P(X = 1) = \binom{9}{1} (0.4458)^1 (0.5542)^8\)

\(P(X = 2) = \binom{9}{2} (0.4458)^2 (0.5542)^7\)

Calculating these:

\(P(X = 0) = (0.5542)^9\)

\(P(X = 1) = 9 \times (0.4458) \times (0.5542)^8\)

\(P(X = 2) = 36 \times (0.4458)^2 \times (0.5542)^7\)

Summing these probabilities gives:

\(P(X < 3) = 0.156\)

(i) The probability that a boy chooses Music is 0.15 (since 75% choose Games and 10% choose Drama, leaving 15% for Music). We use the binomial distribution with parameters: number of trials = 6, probability of success (choosing Music) = 0.15.

The probability that fewer than 3 boys choose Music is:

\(P(0, 1, 2) = (0.85)^6 + (0.15)(0.85)^5 \binom{6}{1} + (0.15)^2(0.85)^4 \binom{6}{2}\)

Calculating each term:

\((0.85)^6 = 0.377149515625\)

\((0.15)(0.85)^5 \binom{6}{1} = 0.2679075\)

\((0.15)^2(0.85)^4 \binom{6}{2} = 0.308\)

Summing these probabilities gives \(0.953\).

(ii) The probability that a student is a Drama student is:

\(P(D) = 0.6 \times 0.1 + 0.4 \times 0.55 = 0.28\)

The probability that a Drama student is a boy is:

\(P(B|D) = \frac{P(B \cap D)}{P(D)} = \frac{0.06}{0.28} = 0.2143\)

The probability that at least 1 of the 5 Drama students is a boy is:

\(P(>1) = 1 - P(0) = 1 - (0.7857)^5\)

\(P(0) = (0.7857)^5 = 0.7078\)

Thus, \(P(>1) = 1 - 0.7078 = 0.2922\)

Therefore, the probability is \(0.701\).

(ii) The mean of \(X\) is given by \(E(X) = 14 \times 0.75 = 10.5\). We evaluate the probabilities for \(X = 10\) and \(X = 11\):

\(P(10) = \binom{14}{10} (0.75)^{10} (0.25)^4 = 0.220\)

\(P(11) = \binom{14}{11} (0.75)^{11} (0.25)^3 = 0.240\)

The value of \(X\) with the highest probability is 11.

(iii) We first find the probability that Sue completes more than 11 puzzles in a month:

\(P(>11) = \binom{14}{12} (0.75)^{12} (0.25)^2 + \binom{14}{13} (0.75)^{13} (0.25)^1 + (0.75)^{14} = 0.281\)

Now, we find the probability that in exactly 3 of the next 5 months, Sue completes more than 11 puzzles:

\(P(3) = \binom{5}{3} (0.281)^3 (0.7189)^2 = 0.115\)

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

1. Constant probability of success.
2. Independent trials.
3. Fixed number of trials with only two outcomes (success or failure).

(ii) Let the random variable \(X\) represent the number of days Julie's train is late out of 9 days. \(X\) follows a binomial distribution with parameters \(n = 9\) and \(p = 0.3\).

We need to find \(P(X > 7) + P(X < 2)\).

\(P(X > 7) = P(X = 8) + P(X = 9)\)

\(P(X = 8) = \binom{9}{8} (0.3)^8 (0.7)^1\)

\(P(X = 9) = \binom{9}{9} (0.3)^9\)

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

\(P(X = 0) = \binom{9}{0} (0.3)^0 (0.7)^9\)

\(P(X = 1) = \binom{9}{1} (0.3)^1 (0.7)^8\)

Calculating these probabilities:

\(P(X = 8) = 9 \times (0.3)^8 \times (0.7) = 0.000015\)

\(P(X = 9) = (0.3)^9 = 0.00000019683\)

\(P(X = 0) = (0.7)^9 = 0.0403536\)

\(P(X = 1) = 9 \times (0.3) \times (0.7)^8 = 0.121060821\)

Adding these probabilities:

\(P(X > 7) + P(X < 2) = 0.000015 + 0.00000019683 + 0.0403536 + 0.121060821 = 0.196\)

This problem can be modeled using a binomial distribution where the probability of success (Martin playing games when his friend phones) is 0.25, and the probability of failure is 0.75. We are looking for the probability of exactly 2 successes in 8 trials.

The probability mass function for a binomial distribution is given by:

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

where \(n\) is the number of trials, \(k\) is the number of successes, and \(p\) is the probability of success.

Substituting the given values:

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

Calculating each part:

\(\binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28\)

\((0.25)^2 = 0.0625\)

\((0.75)^6 = 0.17803125\)

Therefore,

\(P(X = 2) = 28 \times 0.0625 \times 0.17803125 = 0.311\)

Let the random variable \(X\) represent the number of buses in the sample of 11 vehicles. Since 16% of the vehicles are buses, the probability of selecting a bus is \(p = 0.16\).

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

Using the binomial probability formula \(P(X = r) = \binom{n}{r} p^r (1-p)^{n-r}\), where \(n = 11\) and \(p = 0.16\):

\(P(0) = \binom{11}{0} (0.16)^0 (0.84)^{11} = (0.84)^{11} = 0.1469\)

\(P(1) = \binom{11}{1} (0.16)^1 (0.84)^{10} = 11 \times 0.16 \times (0.84)^{10} = 0.30782\)

\(P(2) = \binom{11}{2} (0.16)^2 (0.84)^9 = 55 \times (0.16)^2 \times (0.84)^9 = 0.2931\)

Therefore, \(P(X < 3) = 0.1469 + 0.30782 + 0.2931 = 0.748\).

The problem involves a binomial distribution where the probability of success (a resident being in favour) is 0.8, and the number of trials is 20. We need to find the probability that more than 17 residents are in favour, i.e., \(P(X > 17)\).

Using the binomial probability formula:

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

We calculate:

\(P(X = 18) = \binom{20}{18} (0.8)^{18} (0.2)^2\)

\(P(X = 19) = \binom{20}{19} (0.8)^{19} (0.2)^1\)

\(P(X = 20) = \binom{20}{20} (0.8)^{20}\)

Summing these probabilities gives:

\(P(X > 17) = P(X = 18) + P(X = 19) + P(X = 20)\)

Calculating each term:

\(P(X = 18) = \binom{20}{18} (0.8)^{18} (0.2)^2 = 0.13691\)

\(P(X = 19) = \binom{20}{19} (0.8)^{19} (0.2)^1 = 0.05765\)

\(P(X = 20) = \binom{20}{20} (0.8)^{20} = 0.01153\)

Adding these gives:

\(P(X > 17) = 0.13691 + 0.05765 + 0.01153 = 0.20609\)

Rounding to three decimal places, the probability is 0.206.

The probability of rolling an odd number (1, 3, or 5) is given by:

\(P(\text{odd}) = \frac{1}{6} + \frac{1}{6} + \frac{2}{6} = \frac{2}{3}\)

We need to find the probability of getting at least 7 odd numbers in 8 throws, which is \(P(7) + P(8)\).

Using the binomial probability formula:

\(P(7) = \binom{8}{7} \left( \frac{2}{3} \right)^7 \left( \frac{1}{3} \right)^1 = 8 \times \left( \frac{2}{3} \right)^7 \times \frac{1}{3} = 0.156\)

\(P(8) = \binom{8}{8} \left( \frac{2}{3} \right)^8 = \left( \frac{2}{3} \right)^8 = 0.0390\)

Thus, the probability of obtaining at least 7 odd numbers is:

\(P(7 \text{ or } 8) = 0.156 + 0.0390 = 0.195\)

The probability of throwing a 5 is given as 0.75. Therefore, the probability of not throwing a 5 is:

\(1 - 0.75 = 0.25\)

We need to find the probability of getting at least 8 fives in 10 throws. This is a binomial probability problem where:

\(n = 10, \ p = 0.75\)

The probability of getting exactly \(r\) fives is given by the binomial formula:

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

We need to calculate \(P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)\).

Calculating each term:

\(P(X = 8) = \binom{10}{8} (0.75)^8 (0.25)^2\)

\(P(X = 9) = \binom{10}{9} (0.75)^9 (0.25)^1\)

\(P(X = 10) = \binom{10}{10} (0.75)^{10}\)

Substituting the values:

\(P(X = 8) = 45 \times (0.75)^8 \times (0.25)^2\)

\(P(X = 9) = 10 \times (0.75)^9 \times (0.25)\)

\(P(X = 10) = 1 \times (0.75)^{10}\)

Summing these probabilities gives:

\(P(X \geq 8) = 0.526\)

Let \(X\) be the number of years out of 15 that have New Year's Day on a Saturday. \(X\) follows a binomial distribution with parameters \(n = 15\) and \(p = \frac{1}{7}\).

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 \(P(X = 2)\):

\(P(X = 0) = \left( \frac{6}{7} \right)^{15}\)

\(P(X = 1) = \binom{15}{1} \left( \frac{1}{7} \right) \left( \frac{6}{7} \right)^{14}\)

\(P(X = 2) = \binom{15}{2} \left( \frac{1}{7} \right)^2 \left( \frac{6}{7} \right)^{13}\)

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

\(P(X \leq 2) = \left( \frac{6}{7} \right)^{15} + 15 \left( \frac{1}{7} \right) \left( \frac{6}{7} \right)^{14} + \binom{15}{2} \left( \frac{1}{7} \right)^2 \left( \frac{6}{7} \right)^{13}\)

\(P(X \leq 2) \approx 0.0990 + 0.2476 + 0.2889 = 0.6355\)

Therefore, \(P(X \geq 3) = 1 - 0.6355 = 0.3645\).

Thus, the probability that at least 3 of these years have New Year's Day on a Saturday is approximately 0.365 (accept 0.364).

To solve this problem, we use the binomial probability formula:

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

where \(n\) is the total number of trials, \(k\) is the number of successful trials, and \(p\) is the probability of success on a single trial.

(i) For equal numbers of large and small bands, we need 10 large and 10 small bands. The probability of a band being large is 0.4, and small is 0.6. Thus,

\(P(10 \text{ large and } 10 \text{ small}) = \binom{20}{10} (0.6)^{10} (0.4)^{10}\)

Calculating this gives:

\(= 0.117\)

(ii) For more than 17 small bands, we consider 18, 19, or 20 small bands. The probability of a band being small is 0.6. Thus,

\(P(18, 19, 20 \text{ small}) = \binom{20}{18} (0.6)^{18} (0.4)^{2} + \binom{20}{19} (0.6)^{19} (0.4)^{1} + (0.6)^{20}\)

Calculating each term:

\(= 0.003087 + 0.000487 + 0.00003635\)

Summing these gives:

\(= 0.00361\)

Let the probability that an adult does not wear a watch be 0.09. We are dealing with a binomial distribution where the number of trials is 14 and the probability of success (an adult not wearing a watch) is 0.09.

We need to find the probability that more than 2 adults do not wear a watch, which is equivalent to finding:

\(P(X > 2) = 1 - P(X \leq 2)\)

Calculate \(P(X \leq 2)\) as follows:

\(P(X = 0) = (0.91)^{14}\)

\(P(X = 1) = \binom{14}{1} (0.09)^1 (0.91)^{13}\)

\(P(X = 2) = \binom{14}{2} (0.09)^2 (0.91)^{12}\)

Thus,

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

Substitute the values:

\(P(X = 0) = 0.2670\)

\(P(X = 1) = 0.3698\)

\(P(X = 2) = 0.2377\)

Therefore,

\(P(X \leq 2) = 0.2670 + 0.3698 + 0.2377 = 0.8745\)

Finally,

\(P(X > 2) = 1 - 0.8745 = 0.1255\)

Rounding to three decimal places, the probability is 0.126.

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

1. There are a fixed number of trials.
2. Each trial is independent.
3. There are only two possible outcomes for each trial.
4. The probability of success is constant for each trial.

(ii) Let the random variable \(X\) represent the number of cars made by Ford. \(X\) follows a binomial distribution with parameters \(n = 14\) and \(p = 0.28\).

We need to find \(P(X < 4)\), which is the sum of probabilities \(P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)\).

Using the binomial probability formula:

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

Calculate each probability:

\(P(X = 0) = \binom{14}{0} (0.28)^0 (0.72)^{14} = 0.72^{14} = 0.0101\)

\(P(X = 1) = \binom{14}{1} (0.28)^1 (0.72)^{13} = 14 \times 0.28 \times 0.72^{13} = 0.0548\)

\(P(X = 2) = \binom{14}{2} (0.28)^2 (0.72)^{12} = 91 \times 0.28^2 \times 0.72^{12} = 0.1385\)

\(P(X = 3) = \binom{14}{3} (0.28)^3 (0.72)^{11} = 364 \times 0.28^3 \times 0.72^{11} = 0.2154\)

Summing these probabilities gives:

\(P(X < 4) = 0.0101 + 0.0548 + 0.1385 + 0.2154 = 0.419\)

Let the random variable \(X\) represent the number of damaged tapes in a sample of 15. The probability of a tape being damaged is \(p = 0.2\), and the probability of a tape not being damaged is \(1 - p = 0.8\).

We use the binomial probability formula \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\) where \(n = 15\).

Calculate \(P(X = 0)\):

\(P(0) = (0.8)^{15} = 0.03518\)

Calculate \(P(X = 1)\):

\(P(1) = \binom{15}{1} (0.2)^1 (0.8)^{14} = 15 \times 0.2 \times (0.8)^{14} = 0.1319\)

Calculate \(P(X = 2)\):

\(P(2) = \binom{15}{2} (0.2)^2 (0.8)^{13} = 105 \times (0.2)^2 \times (0.8)^{13} = 0.2309\)

The probability that at most 2 tapes are damaged is:

\(P(X \leq 2) = P(0) + P(1) + P(2) = 0.03518 + 0.1319 + 0.2309 = 0.398\)