Past Exam Questions

The probability that Sue completes all n puzzles correctly is given by \((0.75)^n\).

We need to find the smallest n such that \((0.75)^n < 0.06\).

Taking logarithms on both sides, we have:

\(\log((0.75)^n) < \log(0.06)\)

\(n \cdot \log(0.75) < \log(0.06)\)

\(n > \frac{\log(0.06)}{\log(0.75)}\)

Calculating the right-hand side gives \(n > 9.78\).

Since n must be an integer, the smallest possible value is \(n = 10\).

(i) The probability of not getting a 3 on a single die is \(\frac{5}{6}\). Therefore, the probability of not getting a 3 on all 9 dice is \(\left( \frac{5}{6} \right)^9\). The probability of getting at least one 3 is:

\(1 - \left( \frac{5}{6} \right)^9 = 1 - 0.1938 = 0.806\)

(ii) The probability of getting at least one 3 when \(n\) dice are thrown is \(1 - \left( \frac{5}{6} \right)^n\). We need this probability to be greater than 0.9:

\(1 - \left( \frac{5}{6} \right)^n > 0.9\)

\(\left( \frac{5}{6} \right)^n < 0.1\)

Taking logarithms on both sides:

\(n \log \left( \frac{5}{6} \right) < \log(0.1)\)

\(n > \frac{\log(0.1)}{\log \left( \frac{5}{6} \right)}\)

Calculating gives \(n > 12.6\), so the smallest integer \(n\) is 13.

(i) The probability of performing the routine correctly is 0.65, and incorrectly is 0.35. We use the binomial probability formula:

\(P(X = 5) = (0.65)^5 \times (0.35)^2 \times \binom{7}{5}\)

\(= (0.65)^5 \times (0.35)^2 \times 21\)

\(= 0.298\)

(iii) The expected number of correct performances is given by \(np\), where \(p = 0.65\). We need \(np \geq 8\).

\(0.65n \geq 8\)

\(n \geq \frac{8}{0.65} \approx 12.31\)

The smallest integer \(n\) is 13.

The probability that a tape is not damaged is 0.8 (since 1 in 5 are damaged, so 4 in 5 are not).

The probability that all n tapes in a sample are not damaged is given by (0.8)^n.

The probability that at least one tape is damaged is therefore 1 - (0.8)^n.

We need this probability to be at least 0.85:

1 - (0.8)^n \geq 0.85

(0.8)^n \leq 0.15

\(By trial and error or using logarithms, we find that n = 9 satisfies this inequality.\)

The probability that Janice does not buy an item online in any week is given by:

\(1 - 0.35 = 0.65\)

The probability that Janice does not buy any item online in n weeks is:

\((0.65)^n\)

The probability that Janice buys at least one item online in n weeks is greater than 0.99, so:

\(1 - (0.65)^n > 0.99\)

Rearranging gives:

\((0.65)^n < 0.01\)

Taking logarithms on both sides:

\(n \log(0.65) < \log(0.01)\)

Solving for n gives:

\(n > \frac{\log(0.01)}{\log(0.65)} \approx 10.69\)

Since n must be an integer, the smallest possible value of n is 11.

\(The probability that a customer does not shop online is 0.65 (since 1 - 0.35 = 0.65).\)

The probability that none of the n customers shop online is given by:

(0.65)^n

We want the probability that at least one customer shops online to be greater than 0.95:

1 - (0.65)^n > 0.95

Rearranging gives:

(0.65)^n < 0.05

Taking logarithms on both sides:

n \log(0.65) < \log(0.05)

Solving for n gives:

\(n > \frac{\log(0.05)}{\log(0.65)} \approx 6.95\)

Therefore, the least integer value of n is 7.

The probability that a phone is not made by Company B is 0.65 (since 35% are made by Company B).

The probability that none of the n phones is made by Company B is given by \((0.65)^n\).

We want the probability that at least one phone is made by Company B to be more than 0.98, so:

\(1 - (0.65)^n > 0.98\)

\((0.65)^n < 0.02\)

Taking logarithms on both sides:

\(n \log(0.65) < \log(0.02)\)

\(n > \frac{\log(0.02)}{\log(0.65)}\)

Calculating the right-hand side gives \(n > 9.08\).

Since n must be an integer, the smallest possible value is \(n = 10\).

The probability that a student does not own a car is \(1 - \frac{1}{4} = \frac{3}{4} = 0.75\).

The probability that none of the \(n\) students owns a car is \(0.75^n\).

We need the probability that at least one student owns a car to be greater than 0.995:

\(1 - 0.75^n > 0.995\)

\(0.75^n < 0.005\)

Taking logarithms on both sides:

\(n \log(0.75) < \log(0.005)\)

\(n > \frac{\log(0.005)}{\log(0.75)}\)

Calculating the right-hand side gives \(n > 18.4\).

Therefore, the least possible integer value of \(n\) is 19.

(i) The mean number of cracked eggs is 1.4, so the probability of a cracked egg, p, is given by:

\(p = \frac{1.4}{20} = 0.07\)

The probability of exactly 2 cracked eggs in a box of 20 is calculated using the binomial distribution:

\(P(2) = \binom{20}{2} (0.07)^2 (0.93)^{18}\)

Calculating this gives:

\(P(2) = 0.252\)

(ii) The probability of at least 1 cracked egg is:

\(P(\text{at least 1 cracked egg}) = 1 - P(0)\)

Where:

\(P(0) = (0.93)^{20}\)

Thus:

\(P(\text{at least 1 cracked egg}) = 1 - (0.93)^{20} = 0.766\)

(iii) We need to find the smallest n such that:

\((0.766)^n < 0.01\)

Taking logarithms:

\(n \log(0.766) < \log(0.01)\)

Solving for n gives:

\(n > \frac{\log(0.01)}{\log(0.766)} \approx 17.8\)

Thus, the smallest integer n is 18.

The probability that a person takes more than t minutes is 0.12. Therefore, the probability that a person takes t minutes or less is 0.88.

The probability that none of the n people takes more than t minutes is given by:

\((0.88)^n < 0.003\)

Taking logarithms on both sides, we have:

\(n \log(0.88) < \log(0.003)\)

Solving for n gives:

\(n > \frac{\log(0.003)}{\log(0.88)}\)

Calculating the right-hand side:

\(n > 45.4\)

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

\(n = 46\)

(i) The probability that a person buys something is 0.66 (since 34% do not buy anything). We model this situation with a binomial distribution: \(X \sim B(15, 0.66)\).

The probability that at least 14 people buy something is \(P(X \geq 14) = P(X = 14) + P(X = 15)\).

Using the binomial probability formula:

\(P(X = 14) = \binom{15}{14} (0.66)^{14} (0.34)^1\)

\(P(X = 15) = \binom{15}{15} (0.66)^{15}\)

Calculating these gives:

\(P(X = 14) = 15 \times (0.66)^{14} \times 0.34\)

\(P(X = 15) = (0.66)^{15}\)

Adding these probabilities: \(P(X \geq 14) = 0.0171\).

(ii) The probability that a person spends less than $50 is 0.87. We want \((0.87)^n < 0.04\).

Taking logarithms: \(n \log(0.87) < \log(0.04)\).

Solving for \(n\):

\(n > \frac{\log(0.04)}{\log(0.87)} \approx 23.5\).

The smallest integer \(n\) is 24.

The probability that a person takes a holiday in their own country is 0.65. For a group of n people, the probability that all take a holiday in their own country is given by:

\((0.65)^n < 0.002\)

Taking logarithms on both sides, we have:

\(n \cdot \log(0.65) < \log(0.002)\)

Solving for n, we get:

\(n > \frac{\log(0.002)}{\log(0.65)}\)

Calculating the right-hand side:

\(\log(0.002) \approx -2.69897\)

\(\log(0.65) \approx -0.18709\)

Thus:

\(n > \frac{-2.69897}{-0.18709} \approx 14.43\)

Since n must be an integer, the smallest possible value of n is 15.

(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.