Past Exam Questions

Let the probability of getting the required picture be 0.05 (since there is 1 desired picture out of 20).

1. The probability that Richard does not complete his collection is calculated by not getting the required picture in all 5 attempts:

   \((0.95)^5 = 0.774\)

2. The probability that he has the required picture exactly once is calculated using the binomial probability formula:

   \((0.95)^4 \times (0.05)^1 \times \binom{5}{1} = 0.204\)

3. The probability that he completes his collection with the third picture he looks at is calculated by not getting the picture in the first two attempts and getting it on the third:

   \((0.95)^2 \times (0.05) = 0.0451\)

Let the probability of a bag being underweight be \(p = 0.1\). The number of underweight bags follows a binomial distribution \(X \sim B(10, 0.1)\).

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

Using the binomial probability formula:

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

Calculate each probability:

\(P(X = 0) = \binom{10}{0} (0.1)^0 (0.9)^{10} = 0.348678\)

\(P(X = 1) = \binom{10}{1} (0.1)^1 (0.9)^9 = 0.38742\)

\(P(X = 2) = \binom{10}{2} (0.1)^2 (0.9)^8 = 0.19371\)

Sum these probabilities:

\(P(X < 3) = 0.348678 + 0.38742 + 0.19371 = 0.929808\)

Rounding to three decimal places, the probability is \(0.930\).

Let the random variable \(X\) represent the number of callers connected immediately. \(X\) follows a binomial distribution with parameters \(n = 12\) and \(p = 0.9\).

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

\(P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12)\).

Using the binomial probability formula:

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

Calculate each probability:

\(P(X = 10) = \binom{12}{10} (0.9)^{10} (0.1)^2 = 0.230128\)

\(P(X = 11) = \binom{12}{11} (0.9)^{11} (0.1)^1 = 0.376573\)

\(P(X = 12) = \binom{12}{12} (0.9)^{12} (0.1)^0 = 0.282430\)

\(P(X \geq 10) = 0.230128 + 0.376573 + 0.282430 = 0.889131\)

Thus, \(P(X < 10) = 1 - 0.889131 = 0.110869\).

Therefore, the probability that fewer than 10 callers are connected immediately is approximately 0.111.

(i) Let the probability of throwing a 4 be denoted by \(p\). The expected number of 4s in 30 throws is given by \(30p\). We are given that this average is 6.21, so:

\(30p = 6.21\)

Solving for \(p\), we get:

\(p = \frac{6.21}{30} = 0.207\)

(ii) The variance of a binomial distribution is given by \(np(1-p)\). Here, \(n = 30\) and \(p = 0.207\), so:

\(\text{Variance} = 30 \times 0.207 \times (1 - 0.207)\)

\(= 30 \times 0.207 \times 0.793\)

\(= 4.92\)

(iii) For 15 throws, the probability of obtaining 2 or more 4s can be found using the complement rule. First, find the probability of obtaining fewer than 2 4s (i.e., 0 or 1 4s) using the binomial probability formula:

\(P(X = 0) = \binom{15}{0} (0.207)^0 (0.793)^{15} = (0.793)^{15}\)

\(P(X = 1) = \binom{15}{1} (0.207)^1 (0.793)^{14} = 15 \times 0.207 \times (0.793)^{14}\)

Calculate these probabilities and sum them:

\(P(X < 2) = (0.793)^{15} + 15 \times 0.207 \times (0.793)^{14}\)

Finally, the probability of obtaining 2 or more 4s is:

\(P(X \geq 2) = 1 - P(X < 2) = 0.848\)

(i) The probability of selecting a non-orange disc is \(\frac{200}{300} = \frac{2}{3}\). Since the selection is with replacement, the probability that no orange discs are selected in 5 draws is \(\left( \frac{2}{3} \right)^5\).

(ii) The probability of selecting a disc with a number ending in 6 is \(\frac{10}{100} = \frac{1}{10}\). Using the binomial distribution, the probability of selecting exactly 2 such discs in 5 draws is \(\binom{5}{2} \left( \frac{1}{10} \right)^2 \left( \frac{9}{10} \right)^3\).

(iii) The probability of selecting an orange disc with a number ending in 6 is \(\frac{10}{300} = \frac{1}{30}\). Using the binomial distribution, the probability of selecting exactly 2 such discs in 5 draws is \(\binom{5}{2} \left( \frac{1}{30} \right)^2 \left( \frac{29}{30} \right)^3\).

(iv) The probability of selecting a pink disc is \(\frac{100}{300} = \frac{1}{3}\). For a binomial distribution with \(n = 5\) and \(p = \frac{1}{3}\), the mean is \(np = 5 \times \frac{1}{3} = \frac{5}{3}\) and the variance is \(np(1-p) = 5 \times \frac{1}{3} \times \frac{2}{3} = \frac{10}{9}\).

Let the number of plants per box be denoted by \(n\), and the probability of a plant producing yellow flowers be \(p\). The number of plants producing yellow flowers follows a binomial distribution \(B(n, p)\).

The mean of a binomial distribution is given by \(np = 11\) and the variance is \(np(1-p) = 4.95\).

From \(np = 11\), we have \(p = \frac{11}{n}\).

Substitute \(p\) into the variance equation:

\(np(1-p) = 4.95\)

\(n \left(\frac{11}{n}\right) \left(1 - \frac{11}{n}\right) = 4.95\)

\(11 \left(1 - \frac{11}{n}\right) = 4.95\)

\(11 - \frac{121}{n} = 4.95\)

\(11 - 4.95 = \frac{121}{n}\)

\(6.05 = \frac{121}{n}\)

\(n = \frac{121}{6.05} \approx 20\)

Thus, there are 20 plants per box.

For part (b), we need to find the probability that exactly 12 plants produce yellow flowers, i.e., \(P(X = 12)\).

Using \(n = 20\) and \(p = \frac{11}{20} = 0.55\), the probability mass function of a binomial distribution is:

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

\(P(X = 12) = \binom{20}{12} (0.55)^{12} (0.45)^{8}\)

\(P(X = 12) \approx 0.162\)

Let the number of days Khalid rides his bicycle be a binomial random variable with parameters 45 and 0.6. The variance is given by:

\(\text{Var}(B) = 45 \times 0.6 \times 0.4\)

Similarly, let the number of days Khalid rides his skateboard be a binomial random variable with parameters 45 and 0.4. The variance is given by:

\(\text{Var}(S) = 45 \times 0.4 \times 0.6\)

Since \(\text{Var}(B) = \text{Var}(S)\), the variances are the same.

(i) The probability that a visitor sees all the animals is the product of the individual probabilities:

\(0.9 \times 0.95 \times 0.85 \times 0.1 = 0.0727\).

(ii) Let the random variable \(X\) be the number of visitors who see lions. \(X\) follows a binomial distribution \(X \sim B(12, 0.1)\). We need to find \(P(X < 3)\), which is \(P(0) + P(1) + P(2)\).

\(P(0) = (0.9)^{12}\)

\(P(1) = \binom{12}{1} (0.1)(0.9)^{11}\)

\(P(2) = \binom{12}{2} (0.1)^2(0.9)^{10}\)

Thus, \(P(X < 3) = (0.9)^{12} + \binom{12}{1} (0.1)(0.9)^{11} + \binom{12}{2} (0.1)^2(0.9)^{10} = 0.889\).

(iii) Let \(Y\) be the number of people who see zebras. \(Y\) follows a binomial distribution \(Y \sim B(50, 0.85)\). The mean \(E(Y)\) is given by \(np\) and the variance \(Var(Y)\) is given by \(np(1-p)\).

Mean: \(50 \times 0.85 = 42.5\)

Variance: \(50 \times 0.85 \times 0.15 = 6.375\).

(i) Let the probability of a screw being faulty be \(p\). Given the mean number of faulty screws per packet is 1.2, we have:

\(1.2 = 15p\)

Solving for \(p\), we get:

\(p = \frac{1.2}{15} = 0.08\)

The variance \(\text{Var} = npq\), where \(q = 1 - p\).

\(\text{Var} = 15 \times 0.08 \times 0.92 = 1.104\)

(ii) The probability that a packet contains at most 2 faulty screws is:

\(P(0, 1, 2) = (0.92)^{15} + 15C_1(0.08)(0.92)^{14} + 15C_2(0.08)^2(0.92)^{13}\)

Calculating each term:

\((0.92)^{15} = 0.275\)

\(15 \times 0.08 \times (0.92)^{14} = 0.359\)

\(15C_2 \times (0.08)^2 \times (0.92)^{13} = 0.253\)

Adding these probabilities gives:

\(P(0, 1, 2) = 0.275 + 0.359 + 0.253 = 0.887\)

(iii) The probability that there is at least 1 faulty screw in a packet is:

\(P(\text{at least 1 faulty screw}) = 1 - P(0) = 1 - (0.92)^{15}\)

\(= 1 - 0.275 = 0.7137\)

The probability that there are exactly 7 packets with at least 1 faulty screw is:

\(P(\text{exactly 7 packets}) = 8C_7(0.7137)^7(0.2863)\)

\(= 0.216\)

(i) Let the probability that a random integer is less than or equal to 4 be p. Since there are 4 integers (1, 2, 3, 4) less than or equal to 4 out of 9, p = 4/9. The probability that at least 2 of the 5 integers are less than or equal to 4 is given by:

\(P(\text{at least 2}) = 1 - P(0, 1)\)

\(= 1 - \left(\frac{5}{9}\right)^5 - \binom{5}{1} \left(\frac{4}{9}\right) \left(\frac{5}{9}\right)^4\)

\(= 0.735\)

(ii) We know that the mean \(np = 96\) and the variance \(npq = 32\), where \(q = 1 - p\).

From \(np = 96\), we have \(p = \frac{96}{n}\).

From \(npq = 32\), we have \(np(1-p) = 32\).

Substituting \(p = \frac{96}{n}\) into \(npq = 32\), we get:

\(n \left(\frac{96}{n}\right) \left(1 - \frac{96}{n}\right) = 32\)

\(96 \left(1 - \frac{96}{n}\right) = 32\)

\(96 - \frac{9216}{n} = 32\)

\(\frac{9216}{n} = 64\)

\(n = 144\)

Since \(p = \frac{96}{144} = \frac{2}{3}\), \(k\) is such that \(P(\leq k) = \frac{2}{3}\). Therefore, \(k = 6\).

The average number of 5s in 20 throws is given as 4.8. This implies that the expected value for the number of 5s is 4.8, which can be expressed as:

\(20p = 4.8\)

Solving for \(p\), we get:

\(p = \frac{4.8}{20} = 0.24\)

We need to find the probability that the number of 5s is less than 3 in 20 throws. This is a binomial distribution problem where \(n = 20\) and \(p = 0.24\).

The probability of getting exactly \(k\) successes in a binomial distribution is given by:

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

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

Calculating each term:

\(P(X = 0) = \binom{20}{0} (0.24)^0 (0.76)^{20} = (0.76)^{20}\)

\(P(X = 1) = \binom{20}{1} (0.24)^1 (0.76)^{19} = 20 \times 0.24 \times (0.76)^{19}\)

\(P(X = 2) = \binom{20}{2} (0.24)^2 (0.76)^{18} = \frac{20 \times 19}{2} \times (0.24)^2 \times (0.76)^{18}\)

Summing these probabilities:

\(P(X < 3) = (0.76)^{20} + 20 \times 0.24 \times (0.76)^{19} + \frac{20 \times 19}{2} \times (0.24)^2 \times (0.76)^{18}\)

Calculating the above gives:

\(P(X < 3) = 0.109\)

Let the probability of a biscuit being broken be denoted by \(p\). Since the mean number of broken biscuits is 2.7, we have:

\(18p = 2.7\)

Solving for \(p\), we get:

\(p = \frac{2.7}{18} = 0.15\)

We need to find the probability that the number of broken biscuits \(X\) is between 2 and 4 inclusive, i.e., \(P(2 \leq X \leq 4)\).

\(X\) follows a binomial distribution \(B(18, 0.15)\). Therefore,

\(P(2 \leq X \leq 4) = P(X = 2) + P(X = 3) + P(X = 4)\)

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

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

\(P(X = 3) = \binom{18}{3} (0.15)^3 (0.85)^{15}\)

\(P(X = 4) = \binom{18}{4} (0.15)^4 (0.85)^{14}\)

Summing these probabilities gives:

\(P(2 \leq X \leq 4) = \binom{18}{2} (0.15)^2 (0.85)^{16} + \binom{18}{3} (0.15)^3 (0.85)^{15} + \binom{18}{4} (0.15)^4 (0.85)^{14}\)

Calculating each term:

\(\binom{18}{2} (0.15)^2 (0.85)^{16} \approx 0.231\)

\(\binom{18}{3} (0.15)^3 (0.85)^{15} \approx 0.287\)

\(\binom{18}{4} (0.15)^4 (0.85)^{14} \approx 0.137\)

Adding these gives:

\(P(2 \leq X \leq 4) = 0.231 + 0.287 + 0.137 = 0.655\)

(i) We use the binomial distribution where the probability of going to the park on any day is 0.6. We need to find the probability of going to the park more than 5 times in 7 days, i.e., 6 or 7 times.

The probability of going exactly 6 times is given by:

\(\binom{7}{6} (0.6)^6 (0.4)^1 = 7 \times 0.046656 \times 0.4 = 0.1306\)

The probability of going exactly 7 times is given by:

\(\binom{7}{7} (0.6)^7 = 1 \times 0.279936 = 0.02799\)

Thus, the probability of going more than 5 times is:

\(0.1306 + 0.02799 = 0.159\)

(ii) The variance of a binomial distribution is given by \(np(1-p)\).

Here, \(n = 30\) and \(p = 0.6\).

So, the variance is:

\(30 \times 0.6 \times 0.4 = 7.2\)

Let the number of defective batteries in a pack be a binomial random variable, denoted by \(X\), with parameters \(n = 20\) and probability of defect \(p\).

Given the mean number of defective batteries is 1.6, we have:

\(20p = 1.6\)

Solving for \(p\), we get:

\(p = \frac{1.6}{20} = 0.08\)

We need to find \(P(X > 2)\). Using the complement rule:

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

Calculate \(P(X \leq 2)\) using the binomial probability formula:

\(P(X = 0) = \binom{20}{0} (0.08)^0 (0.92)^{20} = (0.92)^{20}\)

\(P(X = 1) = \binom{20}{1} (0.08)^1 (0.92)^{19} = 20 \times 0.08 \times (0.92)^{19}\)

\(P(X = 2) = \binom{20}{2} (0.08)^2 (0.92)^{18} = \frac{20 \times 19}{2} \times (0.08)^2 \times (0.92)^{18}\)

Thus,

\(P(X \leq 2) = (0.92)^{20} + 20 \times 0.08 \times (0.92)^{19} + \frac{20 \times 19}{2} \times (0.08)^2 \times (0.92)^{18}\)

Substituting the values:

\(P(X \leq 2) = 0.1887 + 0.3281 + 0.2711 = 0.7879\)

Therefore,

\(P(X > 2) = 1 - 0.7879 = 0.212\)

The probability that a customer rates the logo as good is 0.42. Therefore, the probability that a customer does not rate the logo as good is 0.58.

The probability that none of the n customers rate the logo as good is given by:

\((0.58)^n\)

We want the probability that at least one person rates the logo as good to be greater than 0.995. Therefore, we have:

\(1 - (0.58)^n > 0.995\)

Rearranging gives:

\((0.58)^n < 0.005\)

Taking logarithms on both sides:

\(n \log(0.58) < \log(0.005)\)

Solving for n gives:

\(n > \frac{\log(0.005)}{\log(0.58)}\)

Calculating the right-hand side:

\(n > 9.727\)

Since n must be an integer, the smallest value of n is 10.

The probability that a water pistol works properly is 92% or 0.92.

The probability that all n water pistols work properly is given by:

\(P(0) = (0.92)^n\)

We want the probability that at least one does not work properly to be greater than 0.9:

\(1 - P(0) > 0.9\)

Substituting for \(P(0)\):

\(1 - (0.92)^n > 0.9\)

Rearranging gives:

\((0.92)^n < 0.1\)

Taking logarithms on both sides:

\(n \log(0.92) < \log(0.1)\)

Solving for \(n\):

\(n > \frac{\log(0.1)}{\log(0.92)}\)

Calculating gives:

\(n > 27.6\)

Since \(n\) must be an integer, the smallest possible value is:

28

The probability that a mango is not small is 0.85 (since 15% are small).

The probability that none of the n mangoes is small is given by (0.85)^n.

We need to find the largest n such that (0.85)^n \geq 0.1.

Taking logarithms on both sides, we have:

\(n \leq \frac{\log(0.1)}{\log(0.85)} \approx 14.2\)

\(Since n must be an integer, the largest possible value is n = 14.\)

The probability that a student does not have blue eyes is given by:

\(P( ext{not blue eyes}) = 1 - \frac{1}{5} = \frac{4}{5} = 0.8\)

For n students, the probability that none of them has blue eyes is:

\((0.8)^n\)

We need this probability to be less than 0.001:

\((0.8)^n < 0.001\)

Taking logarithms on both sides:

\(n \log(0.8) < \log(0.001)\)

Solving for n:

\(n > \frac{\log(0.001)}{\log(0.8)}\)

Calculating the values:

\(n > \frac{-3}{-0.09691} \approx 30.9\)

Since n must be an integer, the least possible value of n is 31.

(i) The integers between 7 and 21 that are multiples of 5 are 10, 15, and 20. There are 3 such numbers out of 15 total numbers (7 to 21 inclusive), so the probability \(p\) that a randomly chosen integer is a multiple of 5 is \(\frac{3}{15} = 0.2\). Therefore, \(X \sim \text{Bin}(12, 0.2)\).

(ii) We need to calculate \(P(3 \leq X \leq 5)\). This is given by:

\(P(X = 3, 4, 5) = \binom{12}{3} (0.2)^3 (0.8)^9 + \binom{12}{4} (0.2)^4 (0.8)^8 + \binom{12}{5} (0.2)^5 (0.8)^7\)

Calculating each term:

\(\binom{12}{3} (0.2)^3 (0.8)^9 = 0.23622\)

\(\binom{12}{4} (0.2)^4 (0.8)^8 = 0.13287\)

\(\binom{12}{5} (0.2)^5 (0.8)^7 = 0.05315\)

Adding these probabilities gives \(0.23622 + 0.13287 + 0.05315 = 0.422\).

(iii) We want \(P(X = 0) < 0.01\). For \(X \sim \text{Bin}(n, 0.2)\), \(P(X = 0) = (0.8)^n\). We solve:

\((0.8)^n < 0.01\)

Taking logarithms, \(n \log(0.8) < \log(0.01)\).

\(n > \frac{\log(0.01)}{\log(0.8)} \approx 20.65\)

Thus, the smallest integer \(n\) is 21.

(i) Let the random variable \(X\) represent the number of customers who rate the food as ‘good’. \(X\) follows a binomial distribution with parameters \(n = 12\) and \(p = 0.65\). We need to find \(P(2 < X < 12)\).

\(P(2 < X < 12) = 1 - P(X \leq 2)\)

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

\(P(X = 0) = (0.35)^{12}\)

\(P(X = 1) = \binom{12}{1} (0.65)^1 (0.35)^{11}\)

\(P(X = 2) = \binom{12}{2} (0.65)^2 (0.35)^{10}\)

\(P(X \leq 2) = (0.35)^{12} + \binom{12}{1} (0.65)^1 (0.35)^{11} + \binom{12}{2} (0.65)^2 (0.35)^{10}\)

\(P(X \leq 2) = 0.0065359\)

\(P(2 < X < 12) = 1 - 0.0065359 = 0.993\)

(ii) Let \(Y\) be the number of customers who rate the food as ‘poor’. \(Y\) follows a binomial distribution with parameters \(n\) and \(p = 0.13\). We need \(P(Y \geq 1) > 0.95\).

\(P(Y \geq 1) = 1 - P(Y = 0)\)

\(P(Y = 0) = (0.87)^n\)

\(1 - (0.87)^n > 0.95\)

\((0.87)^n < 0.05\)

Solving \((0.87)^n < 0.05\) gives \(n = 22\).