Past Exam Questions

(a) To find \(P(A \cap B)\), we need to determine the outcomes where the score on the red die is divisible by 3 and the sum of the two scores is at least 9.

The possible scores on the red die that are divisible by 3 are 3 and 6.

\(For red die = 3, the possible outcomes for the blue die to make the sum at least 9 are: (3,6).\)

\(For red die = 6, the possible outcomes for the blue die to make the sum at least 9 are: (6,3), (6,4), (6,5), (6,6).\)

Thus, the favorable outcomes are: (3,6), (6,3), (6,4), (6,5), (6,6).

There are 5 favorable outcomes out of 36 possible outcomes, so \(P(A \cap B) = \frac{5}{36}\).

(b) To determine if \(A\) and \(B\) are independent, we check if \(P(A) \times P(B) = P(A \cap B)\).

\(P(A) = \frac{2}{6} = \frac{1}{3}\) since the red die can be 3 or 6.

\(P(B) = \frac{10}{36} = \frac{5}{18}\) since there are 10 outcomes where the sum is at least 9.

\(P(A) \times P(B) = \frac{1}{3} \times \frac{5}{18} = \frac{5}{54}\).

Since \(\frac{5}{54} \neq \frac{5}{36}\), events \(A\) and \(B\) are not independent.

To determine if the events are independent, we need to check if:

\(P(\text{hockey}) \times P(\text{Amos or Benn}) = P(\text{hockey} \cap (\text{Amos or Benn}))\)

Calculate each probability:

\(P(\text{hockey}) = \frac{220}{500} = 0.44\)

\(P(\text{Amos or Benn}) = \frac{86 + 156}{500} = \frac{242}{500} = 0.484\)

\(P(\text{hockey} \cap (\text{Amos or Benn})) = \frac{32 + 72}{500} = \frac{104}{500} = 0.208\)

Check the independence condition:

\(0.44 \times 0.484 = \frac{1331}{6250} \approx 0.213\)

Since \(0.213 \neq 0.208\), the events are not independent.

(i) The number of male students who do not play the piano is the sum of male students who play the guitar and the flute:

\(78 + 42 = 120.\)

The probability that a randomly chosen student is a male who does not play the piano is:

\(\frac{120}{300} = 0.4\).

(ii) To determine if the events are independent, we check if \(P(\text{male}) \times P(\text{not piano}) = P(\text{male} \cap \text{not piano})\).

The probability that a student is male is:

\(P(\text{male}) = \frac{160}{300}\).

The probability that a student does not play the piano is:

\(P(\text{not piano}) = \frac{225}{300}\).

Therefore,

\(P(\text{male}) \times P(\text{not piano}) = \frac{160}{300} \times \frac{225}{300} = \frac{8}{15} \times \frac{3}{4} = \frac{2}{5}\).

The probability that a student is male and does not play the piano is:

\(P(\text{male} \cap \text{not piano}) = \frac{120}{300} = \frac{2}{5}\).

Since \(P(\text{male}) \times P(\text{not piano}) = P(\text{male} \cap \text{not piano})\), the events are independent.

First, calculate the probability of event S (sum is even). The possible sums when two dice are thrown range from 2 to 12. The even sums are 2, 4, 6, 8, 10, and 12. The number of outcomes for each sum is:

• 2: (1,1)
• 4: (1,3), (2,2), (3,1)
• 6: (1,5), (2,4), (3,3), (4,2), (5,1)
• 8: (2,6), (3,5), (4,4), (5,3), (6,2)
• 10: (4,6), (5,5), (6,4)
• 12: (6,6)

Total even outcomes = 18. Therefore, \(P(S) = \frac{18}{36} = \frac{1}{2}\).

Next, calculate the probability of event T (sum is less than 6 or a multiple of 4). The sums less than 6 are 2, 3, 4, 5, and the multiples of 4 are 4, 8, 12. The number of outcomes for each sum is:

• 2: (1,1)
• 3: (1,2), (2,1)
• 4: (1,3), (2,2), (3,1)
• 5: (1,4), (2,3), (3,2), (4,1)
• 8: (2,6), (3,5), (4,4), (5,3), (6,2)
• 12: (6,6)

Total outcomes for T = 16. Therefore, \(P(T) = \frac{16}{36} = \frac{4}{9}\).

Now, calculate \(P(S \cap T)\) (sum is even and either less than 6 or a multiple of 4). The even sums that satisfy T are 2, 4, 8, 12. The number of outcomes for each sum is:

• 2: (1,1)
• 4: (1,3), (2,2), (3,1)
• 8: (2,6), (3,5), (4,4), (5,3), (6,2)
• 12: (6,6)

Total outcomes for \(S \cap T\) = 10. Therefore, \(P(S \cap T) = \frac{10}{36} = \frac{5}{18}\).

For events S and T to be independent, \(P(S) \times P(T) = P(S \cap T)\). Calculate \(P(S) \times P(T) = \frac{1}{2} \times \frac{4}{9} = \frac{4}{18} = \frac{2}{9}\).

Since \(\frac{2}{9} \neq \frac{5}{18}\), the events S and T are not independent.

(a) To arrange the letters such that no consonant is next to another consonant, we first arrange the vowels A, O, E, A. There are 4 vowels, and they can be arranged in \(\frac{4!}{2!}\) ways due to the repetition of A. This gives \(\frac{4!}{2!} = 12\) ways.

Next, we place the 5 consonants (D, M, N, R, D) in the gaps between and around the vowels. There are 5 gaps (before, between, and after the vowels), and we need to choose 4 of these gaps to place the consonants. The consonants can be arranged in \(\frac{5!}{2!}\) ways due to the repetition of D. This gives \(\frac{5!}{2!} = 60\) ways.

The total number of arrangements is \(12 \times 60 = 720\).

(b) To arrange the letters with an A at each end and the Ds not together, we first fix the As at each end: A _ _ _ _ _ _ _ A. This leaves 7 positions for the remaining letters (D, N, D, R, O, M, E).

First, calculate the total arrangements with A at each end: \(\frac{7!}{2!} = 2520\) ways, due to the repetition of D.

Next, calculate the arrangements where the Ds are together. Treat the two Ds as a single unit, so we have (DD, N, R, O, M, E) to arrange, which is \(6! = 720\) ways.

The number of arrangements where the Ds are not together is \(2520 - 720 = 1800\).

(a) The word TOMORROW consists of 8 letters where T, M, and W are unique, O appears 3 times, and R appears 2 times. The total number of different arrangements is given by:

\(\frac{8!}{2!3!} = \frac{40320}{12} = 3360\)

(b) To find the number of arrangements with R at the beginning and end, and the Os not all together, we first consider the arrangement of the remaining 6 letters (T, O, M, O, W, O). The total arrangements of these 6 letters is:

\(\frac{6!}{3!} = 120\)

Next, we subtract the cases where all Os are together. Treating the three Os as a single unit, we arrange T, M, W, and the O unit:

\(4! = 24\)

Thus, the number of arrangements where the Os are not all together is:

\(120 - 24 = 96\)

(a) The word RELEASED consists of 8 letters where E is repeated twice. The number of different arrangements is given by \(\frac{8!}{2!}\). Calculating this gives \(\frac{40320}{2} = 6720\).

(b) Treat LED as a single unit. This gives us 6 units to arrange: LED, R, E, A, S, E. The number of arrangements is \(\frac{6!}{2!}\) because E is repeated. Calculating this gives \(\frac{720}{2} = 360\).

(c) First, calculate the total number of arrangements where A and D are together. Treat AD as a single unit, giving us 7 units: AD, R, E, L, E, S, E. The number of arrangements is \(\frac{7!}{3!}\) because E is repeated. Calculating this gives \(\frac{5040}{6} = 840\). The probability that A and D are not together is \(1 - \frac{840}{6720} = \frac{3}{4}\) or 0.75.

(a) The word CATERPILLAR consists of 11 letters where C, A, T, E, P, I, L, L, A, R, R are present. The letters A, L, and R are repeated. The formula for permutations of letters with repetitions is given by:

\(\frac{11!}{2!2!2!}\)

Calculating this gives:

\(\frac{11!}{2!2!2!} = \frac{39916800}{8} = 4989600\)

(b) To find the number of arrangements with R at the beginning and end, and the two As not together, we use two methods:

Method 1:

Arrange the 7 letters CTEPILL (excluding the two Rs and two As):

\(\frac{7!}{2!}\)

Calculate the number of ways to place As in non-adjacent places:

\(\binom{8}{2}\)

The total number of arrangements is:

\(\frac{7!}{2!} \times \binom{8}{2} = 2520 \times 28 = 70560\)

Method 2:

Total arrangements with R at the beginning and end:

\(\frac{9!}{2!2!}\)

Arrangements with R at ends and As together:

\(\frac{8!}{2!}\)

With As not together:

\(\frac{9!}{2!2!} - \frac{8!}{2!} = 90720 - 20160 = 70560\)

(a) The total number of ways to arrange the letters in RESERVED is \(\frac{8!}{3!2!} = 3360\) because there are 3 Es and 2 Rs.

To have V as the first letter and E as the last letter, we arrange the remaining 6 letters: R, E, S, E, R, D. The number of ways to arrange these is \(\frac{6!}{2!2!} = 180\).

The probability is \(\frac{180}{3360} = \frac{3}{56}\) or 0.0536.

(b) Consider Rs together and Es together as single units: [RR] and [EEE]. This leaves us with the units [RR], [EEE], S, V, D to arrange, which is 5! ways.

The number of ways to arrange the Es together is \(\frac{6!}{2!} = 360\).

The probability is \(\frac{5!}{\frac{6!}{2!}} = \frac{120}{360} = \frac{1}{3}\).

(a) Consider the 3 Es as a single unit. This gives us the units: S, H, O, P, K, (EEE), P, R. There are 8 units in total. The number of ways to arrange these 8 units is \(\frac{8!}{2!}\) because the Ps are repeated. Therefore, the number of ways is \(\frac{8!}{2!} = 20160\).

(b) The total number of ways to arrange the letters of SHOPKEEPER is \(\frac{10!}{2!3!} = 302400\). If the Ps are together, treat them as a single unit: S, H, O, (PP), K, E, E, E, R. This gives us 9 units. The number of ways to arrange these 9 units is \(\frac{9!}{3!} = 60480\). Therefore, the number of ways the Ps are not together is \(302400 - 60480 = 241920\).

(c) The number of ways to have Es at the beginning and end is the number of ways to arrange the remaining 8 letters: S, H, O, P, K, E, P, R. This is \(\frac{8!}{2!} = 20160\). The total number of arrangements is \(\frac{10!}{2!3!} = 302400\). Therefore, the probability is \(\frac{20160}{302400} = \frac{1}{15}\).

(a) The word CELESTIAL has 9 letters with 2 Es. The number of different arrangements is given by:

\(\frac{9!}{2!} = 90,720\)

(b) Fixing C as the first letter, T as the fifth letter, and E as the last letter, we arrange the remaining 6 letters (E, L, S, I, A, L). The number of arrangements is:

\(\frac{6!}{2!} = 360\)

(c) First, calculate the number of arrangements where the two Es are together. Treat the two Es as a single unit, reducing the problem to arranging 8 units (EE, C, L, S, T, I, A, L):

\(\frac{8!}{2!} = 20,160\)

The number of arrangements where the Es are not together is:

\(90,720 - 20,160 = 70,560\)

The probability that the Es are not together is:

\(\frac{70,560}{90,720} = \frac{7}{9} \approx 0.778\)

(a) To arrange the letters of SUMMERTIME with an E at the beginning and an E at the end, we fix the Es in these positions. This leaves us with 8 letters: S, U, M, M, R, T, I, M. The number of ways to arrange these 8 letters is given by:

\(\frac{8!}{3!}\)

where 3! accounts for the repetition of the letter M. Calculating this gives:

\(\frac{8!}{3!} = \frac{40320}{6} = 6720\)

(b) To find the number of ways to arrange the letters of SUMMERTIME such that the Es are not together, we first find the total number of arrangements of the letters and then subtract the number of arrangements where the Es are together.

The total number of arrangements of the letters is:

\(\frac{10!}{2!3!}\)

where 2! accounts for the repetition of E and 3! for M. Calculating this gives:

\(\frac{10!}{2!3!} = \frac{3628800}{12} = 302400\)

Next, consider the Es as a single unit. This gives us 9 units to arrange: (EE), S, U, M, M, R, T, I, M. The number of ways to arrange these is:

\(\frac{9!}{3!}\)

Calculating this gives:

\(\frac{9!}{3!} = \frac{362880}{6} = 60480\)

Therefore, the number of arrangements where the Es are not together is:

\(302400 - 60480 = 241920\)

(a) Treat the three Es as a single unit and the two Ls as another single unit. This reduces the problem to arranging 6 units: {EEE, LL, J, W, R, Y}. The number of arrangements is given by:

\(\frac{6!}{1!1!1!1!1!1!} = 6! = 720\)

(b) First, calculate the total number of arrangements of the letters in JEWELLERY without any restrictions:

\(\frac{9!}{3!2!} = 30,240\)

Next, calculate the number of arrangements where the two Ls are together. Treat the two Ls as a single unit, reducing the problem to arranging 8 units: {LL, J, E, E, E, W, R, Y}. The number of arrangements is:

\(\frac{8!}{3!} = 6,720\)

Subtract the number of arrangements with the Ls together from the total arrangements to find the number of arrangements where the Ls are not together:

\(30,240 - 6,720 = 23,520\)

(i) The word CORRIDORS has 9 letters where R appears twice. The number of different arrangements is given by:

\(\frac{9!}{2!} = \frac{362880}{2} = 30240\)

(ii) For the arrangements where the first letter is D and the last letter is R:

We fix D at the start and R at the end, leaving 7 letters: O, R, I, D, O, R, S.

The number of arrangements is:

\(\frac{7!}{2!} = \frac{5040}{2} = 2520\)

For the arrangements where the first letter is D and the last letter is O:

We fix D at the start and O at the end, leaving 7 letters: C, O, R, R, I, D, R.

The number of arrangements is:

\(\frac{7!}{3!} = \frac{5040}{6} = 840\)

\(Total arrangements = 1260 + 840 = 2100.\)

(i) Consider the three Os as a single unit and the two Ts as another unit. This reduces the problem to arranging the units: \(\{OOO, TT, A, D, S, L\}\), which is 6 units in total. The number of arrangements is \(6! = 720\).

(ii) First, calculate the total number of arrangements of the letters in TOADSTOOL. There are 9 letters with 2 Ts and 3 Os, so the total arrangements are \(\frac{9!}{2!3!} = 30,240\). Next, calculate the number of arrangements where the Ts are together. Treat the Ts as a single unit, reducing the problem to arranging \(\{TT, O, O, O, A, D, S, L\}\), which is 8 units. The number of arrangements is \(\frac{8!}{3!} = 6,720\). Therefore, the number of arrangements where the Ts are not together is \(30,240 - 6,720 = 23,520\).

(iii) To find the probability that a randomly chosen arrangement has a T at the beginning and a T at the end, consider the remaining 7 letters \(\{O, O, O, A, D, S, L\}\). The number of arrangements of these letters is \(\frac{7!}{3!} = 840\). The probability is then \(\frac{840}{30,240} = \frac{1}{36}\) or 0.0278.

(a) First, calculate the total number of arrangements of the letters in CASABLANCA. The word has 10 letters with the following repetitions: A appears 3 times, C appears 2 times. The total arrangements are given by:

\(\frac{10!}{2!4!} = 75600\)

Next, calculate the number of arrangements where the two Cs are together. Treat the two Cs as a single unit, reducing the problem to arranging 9 units (CC, A, S, A, B, L, A, N, A). The arrangements are:

\(\frac{9!}{4!} = 15120\)

Thus, the number of arrangements where the Cs are not together is:

\(75600 - 15120 = 60480\)

(b) Fix A at the beginning and A at the end, leaving 8 positions to fill. The sequence is A _ _ _ C _ _ _ C _ A. The three letters between the Cs can be chosen from the remaining 6 letters (S, A, B, L, N, A). The number of ways to choose and arrange these 3 letters is:

\(\frac{6!}{2!} \times 4 = 1440\)

(i) Treat the four Es as a single unit. This gives us 9 units to arrange: S, T, EEEE, P, L, C, H, A, S. The number of ways to arrange these 9 units is \(\frac{9!}{2!}\) because there are 2 indistinguishable Ss. Therefore, the number of ways is \(\frac{9!}{2!} = 181,440\).

(ii) First, find the total number of arrangements of the letters in STEEPLECHASE. There are 12 letters with 4 Es and 2 Ss, so the total number of arrangements is \(\frac{12!}{2!4!} = 9,979,200\).

Next, find the number of arrangements where the Ss are together. Treat the two Ss as a single unit, giving us 11 units to arrange: SS, T, E, E, E, E, P, L, C, H, A. The number of ways to arrange these 11 units is \(\frac{11!}{4!} = 1,663,200\).

Finally, subtract the number of arrangements where the Ss are together from the total number of arrangements to find the number of arrangements where the Ss are not together: \(9,979,200 - 1,663,200 = 8,316,000\).

(i) Consider the odd digits (3, 7, 7, 7) as one block and the even digits (2, 2, 8) as another block. The arrangement of these two blocks can be done in 2! ways.

Within the odd block, the digits can be arranged as \(\frac{4!}{3!}\) because there are three 7s.

Within the even block, the digits can be arranged as \(\frac{3!}{2!}\) because there are two 2s.

Therefore, the total number of arrangements is \(2! \times \frac{4!}{3!} \times \frac{3!}{2!} = 2 \times 4 \times 3 = 24\).

(ii) First, calculate the total number of arrangements of the digits without any restriction: \(\frac{7!}{2!3!} = 420\).

Next, calculate the number of arrangements where the two 2s are together. Treat the two 2s as a single unit, so we have 6 units to arrange: \(\frac{6!}{3!} = 120\).

The number of arrangements where the 2s are not together is the total arrangements minus the arrangements where the 2s are together: \(420 - 120 = 300\).

(i) The word CAMERAMAN consists of 9 letters where A appears 3 times and M appears 2 times. The number of arrangements without restrictions is given by:

\(\frac{9!}{3! \times 2!} = \frac{362880}{6 \times 2} = 30240\)

(ii) If the As occupy the 1st, 5th, and 9th positions, we arrange the remaining 6 letters (C, M, E, R, M, N). The number of arrangements is:

\(\frac{6!}{2!} = \frac{720}{2} = 360\)

(iii) If there is exactly one letter between the Ms, consider the block M_X_M where X is any letter. We have 7 positions for this block among the remaining letters (A, A, A, C, E, R, N). The number of arrangements is:

Method 1: Arrange the 7 letters (excluding M) and multiply by 7 positions for the block:

\(\frac{7!}{3!} \times 7 = \frac{5040}{6} \times 7 = 5880\)

Method 2: Choose a letter between Ms and arrange:

\(\frac{6!}{2!} \times 7 + \frac{6!}{3!} \times 7 = 2520 + 3360 = 5880\)

The word MISSISSIPPI consists of 11 letters where the letter M appears 1 time, I appears 4 times, S appears 4 times, and P appears 2 times.

The formula for the number of different arrangements of letters in a word is given by:

\(\frac{n!}{n_1! \times n_2! \times \ldots \times n_k!}\)

where \(n\) is the total number of letters, and \(n_1, n_2, \ldots, n_k\) are the frequencies of the repeated letters.

For MISSISSIPPI:

\(n = 11, \quad n_1 = 4 \text{ (I)}, \quad n_2 = 4 \text{ (S)}, \quad n_3 = 2 \text{ (P)}\)

Substitute these values into the formula:

\(\frac{11!}{4! \times 4! \times 2!}\)

Calculate the factorials:

\(11! = 39916800, \quad 4! = 24, \quad 2! = 2\)

\(\frac{39916800}{24 \times 24 \times 2} = \frac{39916800}{1152} = 34650\)

Therefore, the number of different arrangements is 34650.