The probability of rolling a 1 or 2 is \(\frac{2}{6}\), and the probability of rolling a 3, 4, 5, or 6 is \(\frac{4}{6}\).

For bag A, the probability of picking two white marbles is:

\(\frac{8}{15} \times \frac{7}{14}\)

For bag B, the probability of picking two white marbles is:

\(\frac{6}{15} \times \frac{5}{14}\)

The total probability that both marbles are white is:

\(\frac{2}{6} \times \frac{8}{15} \times \frac{7}{14} + \frac{4}{6} \times \frac{6}{15} \times \frac{5}{14}\)

Calculating each part:

\(\frac{2}{6} \times \frac{8}{15} \times \frac{7}{14} = \frac{56}{630}\)

\(\frac{4}{6} \times \frac{6}{15} \times \frac{5}{14} = \frac{60}{630}\)

Adding these probabilities gives:

\(\frac{56}{630} + \frac{60}{630} = \frac{116}{630} = \frac{58}{315}\)

Thus, the probability that both marbles are white is \(\frac{58}{315}\) or 0.184.

To find the probability that exactly 1 student is studying Music and exactly 2 are boys, we consider two scenarios:

1. 1 boy studying Music, 1 boy not studying Music, and 1 girl not studying Music.
2. 1 girl studying Music, 2 boys not studying Music.

For the first scenario:

\(P(B \text{ M}) \times P(B \text{ NM}) \times P(G \text{ NM}) = \frac{40}{160} \times \frac{56}{159} \times \frac{52}{158} \times 3! = 0.17387\)

For the second scenario:

\(P(G \text{ M}) \times P(B \text{ NM}) \times P(B \text{ NM}) = \frac{12}{160} \times \frac{56}{159} \times \frac{55}{158} \times \frac{3!}{2!} = 0.02759\)

Adding both probabilities gives:

\(P = 0.17387 + 0.02759 = 0.201\)

The possible outcomes for the sum of 9 when the die is thrown twice are: (3, 6), (6, 3), (4, 5), and (5, 4).

The probability of rolling a 3 is 0.1 and the probability of rolling a 6 is 0.3. Thus, the probability of (3, 6) is:

\(P(3, 6) = 0.1 \times 0.3 = 0.03\)

The probability of (6, 3) is the same as (3, 6) because the events are independent:

\(P(6, 3) = 0.3 \times 0.1 = 0.03\)

The probability of rolling a 4 is 0.2 and the probability of rolling a 5 is 0.1. Thus, the probability of (4, 5) is:

\(P(4, 5) = 0.2 \times 0.1 = 0.02\)

The probability of (5, 4) is the same as (4, 5):

\(P(5, 4) = 0.1 \times 0.2 = 0.02\)

Therefore, the total probability that the sum is 9 is:

\(P(\text{sum is 9}) = P(3, 6) + P(6, 3) + P(4, 5) + P(5, 4) = 0.03 + 0.03 + 0.02 + 0.02 = 0.1\)

To find \(r\), note that the only way to get a score of 6 is by landing on 3 on both spinners, so \(P(\text{score is } 6) = r^2 = \frac{1}{36}\). Solving for \(r\), we get \(r = \frac{1}{6}\).

For a score of 5, the possible outcomes are (2, 3) and (3, 2). Thus, \(P(\text{score is } 5) = P(2, 3) + P(3, 2) = qr + rq = 2qr = \frac{1}{9}\). Substituting \(r = \frac{1}{6}\), we have:

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

\(\frac{q}{3} = \frac{1}{9}\)

\(q = \frac{1}{3}\)

Since \(p + q + r = 1\), we have:

\(p + \frac{1}{3} + \frac{1}{6} = 1\)

\(p + \frac{1}{2} = 1\)

\(p = \frac{1}{2}\)

The total number of balloons is 40 (10 pink + 9 yellow + 12 green + 9 white).

We need to find the probability of selecting exactly 3 green balloons out of 7 selected balloons.

Using combinations, the number of ways to choose 3 green balloons from 12 is given by:

\(\binom{12}{3}\)

The number of ways to choose the remaining 4 balloons from the 28 non-green balloons is:

\(\binom{28}{4}\)

The total number of ways to choose 7 balloons from 40 is:

\(\binom{40}{7}\)

The probability is then given by:

\(\frac{\binom{12}{3} \times \binom{28}{4}}{\binom{40}{7}}\)

Calculating each part:

\(\binom{12}{3} = 220\)

\(\binom{28}{4} = 20475\)

\(\binom{40}{7} = 18643560\)

Thus, the probability is:

\(\frac{220 \times 20475}{18643560} = 0.242\)

(i) The probability of getting a green robot from one packet is \(\frac{1}{4}\) since there are four equally likely colours.

(ii) The probability that Nick gets his first green robot on the fifth packet means he does not get a green robot in the first four packets and then gets a green robot on the fifth. The probability of not getting a green robot in one packet is \(\frac{3}{4}\). Therefore, the probability is \(\left( \frac{3}{4} \right)^4 \times \frac{1}{4} = \frac{81}{1024} = 0.0791\).

(iii) The probability that the first four packets Amos opens all contain different coloured robots is calculated by considering the permutations of the four colours. The probability is \(\frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \times 4! = \frac{3}{32}\).

There are 3 even numbers (2, 4, 6) and 2 odd numbers (1, 7) in the box.

The total number of ways to choose 3 discs from 5 is given by \(\binom{5}{3} = 10\).

To find the number of favorable outcomes (2 even and 1 odd), we can choose 2 even numbers from 3 and 1 odd number from 2:

\(\binom{3}{2} \times \binom{2}{1} = 3 \times 2 = 6\).

Thus, the probability is:

\(\frac{6}{10} = 0.6\).

The tree diagram is constructed as follows:

1. First choice: Sharik has a probability of \(\frac{1}{3}\) of choosing the correct answer (C) and \(\frac{2}{3}\) of choosing a wrong answer (W).
2. Second choice (if first was wrong): From the remaining two answers, the probability of choosing the correct answer is \(\frac{1}{2}\) and the wrong answer is \(\frac{1}{2}\).
3. Third choice (if second was also wrong): The remaining answer must be correct, so the probability is 1.

The tree diagram shows these probabilities and the paths leading to the correct answer.

(i) To find the probability that there is a winner after exactly two sets, we consider two scenarios: Roger wins both sets (RR) or Andy wins both sets (AA).

The probability that Roger wins both sets (RR) is:

\(P(RR) = 0.6 \times 0.7 = 0.42\)

The probability that Andy wins both sets (AA) is:

\(P(AA) = 0.4 \times 0.75 = 0.3\)

Thus, the probability that there is a winner after exactly two sets is:

\(P(2 \text{ sets in match}) = P(RR) + P(AA) = 0.42 + 0.3 = 0.72\)

(ii) To find the probability that Andy wins the match given that there is a winner after exactly two sets, we use conditional probability:

\(P(A \text{ wins and 2 sets}) = P(AA) = 0.3\)

\(P(\text{2 sets}) = 0.72\)

Thus, the probability is:

\(\frac{P(AA)}{P(\text{2 sets})} = \frac{0.3}{0.72} = \frac{5}{12}\)

To find the probability that Nur chooses a play-house, we calculate the probability for each playground and sum them up.

For Playground X, the probability of choosing a play-house is:

\(P(X \text{ and } P) = \frac{1}{4} \times \frac{4}{9} = \frac{1}{9}\)

For Playground Y, the probability of choosing a play-house is:

\(P(Y \text{ and } P) = \frac{1}{4} \times \frac{2}{12} = \frac{1}{24}\)

For Playground Z, the probability of choosing a play-house is:

\(P(Z \text{ and } P) = \frac{1}{2} \times \frac{1}{16} = \frac{1}{32}\)

Summing these probabilities gives:

\(P(P) = \frac{1}{9} + \frac{1}{24} + \frac{1}{32} = \frac{53}{288} \approx 0.184\)

(i) To find the probability that the two digits chosen are equal, consider the digits 1, 3, and 5 in the number 113 333 555. The probabilities are:

\(P(1, 1) = \frac{2}{9} \times \frac{1}{8}\)

\(P(3, 3) = \frac{4}{9} \times \frac{3}{8}\)

\(P(5, 5) = \frac{3}{9} \times \frac{2}{8}\)

Summing these probabilities gives:

\(\frac{2}{9} \times \frac{1}{8} + \frac{4}{9} \times \frac{3}{8} + \frac{3}{9} \times \frac{2}{8} = \frac{5}{18}\)

(ii) To find the probability that one digit is a 5 and one digit is not a 5, consider:

\(P(5, \overline{5}) + P(\overline{5}, 5)\)

\(= \frac{3}{9} \times \frac{6}{8} + \frac{6}{9} \times \frac{3}{8} = \frac{36}{72} = \frac{1}{2}\)

(a) The probability that it does not rain on Sunday is \(0.6\). If it does not rain on any day, the probability for each subsequent day is \(0.8\). Therefore, the probability that it does not rain on any of the 4 days is:

\(P(\text{no rain}) = 0.6 \times 0.8^3 = \frac{192}{625}\)

(b) The probability that it does not rain on Sunday is \(0.6\). The probability that it rains on Tuesday, given it did not rain on Monday, is \(0.2\). Therefore, the probability that the first day it rains is Tuesday is:

\(0.6 \times 0.8 \times 0.2 = \frac{12}{125}\)

(c) The probability that it rains on exactly one of the 4 days can be calculated by considering the scenarios where it rains on one specific day and not on the others. The scenarios are:

• Rains on Sunday: \(0.4 \times 0.3 \times 0.8 \times 0.8 = \frac{48}{625}\)
• Rains on Monday: \(0.6 \times 0.2 \times 0.3 \times 0.8 = \frac{18}{625}\)
• Rains on Tuesday: \(0.6 \times 0.8 \times 0.2 \times 0.3 = \frac{18}{625}\)
• Rains on Wednesday: \(0.6 \times 0.8 \times 0.8 \times 0.2 = \frac{48}{625}\)

Adding these probabilities gives:

\(\frac{48}{625} + \frac{18}{625} + \frac{18}{625} + \frac{48}{625} = \frac{132}{625}\)

The probability of picking a robot card from Jack's pack is \(\frac{10}{15} = \frac{2}{3}\).

Emma has 7 robot cards, so the total number of cards in Emma's pack is \(x + 4\) because \(x - 3\) cards are aeroplanes, making \(x - 3 + 7 = x + 4\) cards in total.

The probability of picking a robot card from Emma's pack is \(\frac{7}{x+4}\).

The probability that both cards have pictures of robots is \(\frac{2}{3} \times \frac{7}{x+4} = \frac{7}{18}\).

Setting up the equation: \(\frac{2}{3} \times \frac{7}{x+4} = \frac{7}{18}\).

Solving for \(x\):

\(\frac{14}{3(x+4)} = \frac{7}{18}\)

Cross-multiply: \(14 \times 18 = 7 \times 3(x+4)\)

\(252 = 21(x+4)\)

\(252 = 21x + 84\)

\(168 = 21x\)

\(x = 8\)

Let the houses with numbers greater than 14 be 16, 18, 20, 22, and 24. There are 5 such houses.

We need to find the probability that at least 2 newspapers are delivered to these 5 houses.

This can be calculated as \(P(2) + P(3)\) or \(1 - P(0, 1)\).

Calculate \(P(2)\):

\(P(2) = \frac{5}{12} \times \frac{4}{11} \times \frac{7}{10} \times \binom{3}{2}\)

Calculate \(P(3)\):

\(P(3) = \frac{5}{12} \times \frac{4}{11} \times \frac{3}{10}\)

Thus, \(P(\text{at least 2}) = P(2) + P(3) = \frac{4}{11}\).

Alternatively, using combinations:

\(\frac{\binom{5}{3} + \binom{5}{2} \times \binom{7}{1}}{\binom{12}{3}} = \frac{4}{11}\).

Each tile can be black, white, or grey, giving a probability of \(\frac{1}{3}\) for each color. To ensure no two adjacent tiles are the same color, the first tile can be any color, but each subsequent tile must be a different color from the one before it.

For each of the 7 subsequent tiles, there are 2 choices (not the same as the previous tile), so the probability for each is \(\frac{2}{3}\).

The total probability is:

\(\left( \frac{2}{3} \right)^7 = \frac{128}{2187}\)

Thus, the probability that no two adjacent tiles are the same color is \(\frac{128}{2187}\) or approximately 0.0585.

(i) To complete the table, we use the given information:

• There are 12 biscuits wrapped in gold foil, 7 of which are chocolate-covered. Therefore, 5 are not chocolate-covered.
• There are 17 chocolate-covered biscuits in total, so 10 must be unwrapped.
• There are 30 biscuits in total, so 13 are not chocolate-covered. Of these, 5 are wrapped, so 8 are unwrapped.

(ii) The probability that a biscuit is wrapped in gold foil is given by:

\(\frac{12}{30} = 0.4\)

(iii) The probability that an unwrapped biscuit is chocolate-covered is:

\(\frac{10}{18} = \frac{5}{9} \approx 0.556\)

(iv) The probability that a biscuit is unwrapped, given that it is chocolate-covered, is:

\(\frac{10}{17} \approx 0.588\)

(v) The probability that Nasir takes exactly 2 wrapped biscuits is calculated using combinations:

\(\frac{\binom{12}{2} \times \binom{18}{2}}{\binom{30}{4}}\)

Calculating each part:

• \(\binom{12}{2} = 66\)
• \(\binom{18}{2} = 153\)
• \(\binom{30}{4} = 27,405\)

Thus, the probability is:

\(\frac{66 \times 153}{27,405} = 0.368\)

Let the probability of choosing a rope of length 3 m be \(p = \frac{4}{5}\) and the probability of choosing a rope of length 5 m be \(1 - p = \frac{1}{5}\).

(ii) The probability that two ropes have different lengths is given by:

\(P(3, 5) + P(5, 3) = p \times (1-p) + (1-p) \times p\)

\(= \frac{4}{5} \times \frac{1}{5} + \frac{1}{5} \times \frac{4}{5}\)

\(= \frac{8}{25}\) (0.32)

(iii) The probability that three ropes have a total length of 11 m can occur in the following ways: (3, 3, 5), (3, 5, 3), (5, 3, 3). The probability for each arrangement is:

\(P(3, 3, 5) = p \times p \times (1-p)\)

\(= \frac{4}{5} \times \frac{4}{5} \times \frac{1}{5}\)

\(= \frac{16}{125}\)

Since there are 3 such arrangements, the total probability is:

\(3 \times \frac{16}{125} = \frac{48}{125}\) (0.384)

Let the probability of landing on the blue side be \(p\).

Then the probability of landing on the red side is \(4p\), and the probability of landing on the green side is \(3p\).

Since the total probability must equal 1, we have:

\(4p + p + 3p = 1\)

\(8p = 1\)

\(p = \frac{1}{8}\)

Thus, the probability of landing on the blue side is \(\frac{1}{8}\).

For part (ii), the probabilities are:

\(P(R) = \frac{1}{2}\), \(P(B) = \frac{1}{8}\), \(P(G) = \frac{3}{8}\).

The probability that the spinner lands on a different coloured side each time is:

\(P(\text{all different}) = \frac{1}{2} \times \frac{1}{8} \times \frac{3}{8} \times 3!\)

\(= \frac{9}{64}\)

Therefore, the probability is \(\frac{9}{64}\) or approximately 0.141.

To find the probability that Ronnie wins, we consider the scenarios where Ronnie draws the first yellow ball.

1. Probability that Julie draws a green ball first and Ronnie draws a yellow ball second:

\(P(GY) = \frac{5}{8} \times \frac{3}{7} = \frac{15}{56}\)

2. Probability that both Julie and Ronnie draw green balls first, and Julie draws a green ball again, then Ronnie draws a yellow ball:

\(P(GGGY) = \frac{5}{8} \times \frac{4}{7} \times \frac{3}{6} \times \frac{3}{5} = \frac{3}{28}\)

3. Probability that the first five draws are green balls, and Ronnie draws a yellow ball on his third turn:

\(P(GGGGGY) = \frac{5}{8} \times \frac{4}{7} \times \frac{3}{6} \times \frac{2}{5} \times \frac{1}{4} \times \frac{3}{3} = \frac{1}{56}\)

Adding these probabilities gives the total probability that Ronnie wins:

\(P(R \text{ wins}) = \frac{15}{56} + \frac{3}{28} + \frac{1}{56} = \frac{11}{28}\)

To find the probability that Mrs Lin sits directly behind a student and Mrs Brown sits in the front row, we can use the following approach:

There are 14 seats, and 12 passengers are seated randomly. We need to calculate the probability of the specific seating arrangement.

First, calculate the total number of ways to arrange 12 passengers in 14 seats: \(^{14}P_{12}\).

Next, calculate the favorable outcomes:

1. Choose 3 seats for Mrs Brown in the front row: 3 ways.
2. Choose 10 seats for Mrs Lin to sit directly behind a student: 10 ways.
3. Choose 5 students to sit in front of Mrs Lin: 5 ways.

The number of favorable outcomes is: \(3 \times 10 \times 5 \times ^{11}P_9\).

The probability is then given by: \(\frac{3 \times 10 \times 5 \times ^{11}P_9}{^{14}P_{12}}\).

Calculating this gives: \(\frac{150}{2184} = 0.0687\).

First, determine the probabilities for each person choosing each entrance:

• Rick: \(P(A) = \frac{1}{3}\), \(P(B) = P(C) = P(D) = \frac{2}{9}\)
• Brenda: \(P(A) = P(B) = P(C) = P(D) = \frac{1}{4}\)
• Ali: \(P(C) = \frac{2}{7}\), \(P(D) = \frac{3}{5}\), \(P(A) = P(B) = \frac{2}{35}\)

(i) To find the probability that at least 2 friends choose entrance B:

Calculate the probability for each scenario where at least 2 choose B:

• \(P(\text{Rick B, Brenda B, Ali not B}) = \frac{2}{9} \times \frac{1}{4} \times \left(1 - \frac{2}{35}\right) = \frac{11}{210}\)
• \(P(\text{Rick B, Brenda not B, Ali B}) = \frac{2}{9} \times \left(1 - \frac{1}{4}\right) \times \frac{2}{35} = \frac{2}{210}\)
• \(P(\text{Rick not B, Brenda B, Ali B}) = \left(1 - \frac{2}{9}\right) \times \frac{1}{4} \times \frac{2}{35} = \frac{1}{90}\)
• \(P(\text{Rick B, Brenda B, Ali B}) = \frac{2}{9} \times \frac{1}{4} \times \frac{2}{35} = \frac{1}{315}\)

Sum these probabilities: \(\frac{11}{210} + \frac{2}{210} + \frac{1}{90} + \frac{1}{315} = \frac{24}{315} = \frac{8}{105}\)

(ii) To find the probability that all three choose the same entrance:

• \(P(\text{entrance A}) = \frac{1}{3} \times \frac{1}{4} \times \frac{2}{35} = \frac{1}{210}\)
• \(P(\text{entrance B}) = \frac{2}{9} \times \frac{1}{4} \times \frac{2}{35} = \frac{1}{315}\)
• \(P(\text{entrance C}) = \frac{2}{9} \times \frac{1}{4} \times \frac{2}{7} = \frac{1}{63}\)
• \(P(\text{entrance D}) = \frac{2}{9} \times \frac{1}{4} \times \frac{3}{5} = \frac{1}{30}\)

Sum these probabilities: \(\frac{1}{210} + \frac{1}{315} + \frac{1}{63} + \frac{1}{30} = \frac{2}{35}\)

The probability that the dog barks can be found by considering two scenarios: when they go to the park and when they do not.

1. Probability they go to the park and the dog barks:

\(P(\text{park, bark}) = 0.6 \times 0.35 = 0.21\)

2. Probability they do not go to the park and the dog barks:

\(P(\text{not park, bark}) = 0.4 \times 0.75 = 0.3\)

3. Total probability that the dog barks:

\(P(\text{bark}) = P(\text{park, bark}) + P(\text{not park, bark}) = 0.21 + 0.3 = 0.51\)

Let the probability that it is not raining be \(1 - x\).

The probability that Aran does not wear a hat when it is not raining is \(1 - 0.3 = 0.7\).

The probability that Aran does not wear a scarf when he does not wear a hat is \(1 - 0.1 = 0.9\).

The probability that it is not raining and Aran is not wearing a hat or a scarf is given by:

\((1-x) \times 0.7 \times 0.9 = 0.36\)

Solving for \(x\):

\((1-x) \times 0.63 = 0.36\)

\(0.63 - 0.63x = 0.36\)

\(0.63x = 0.63 - 0.36\)

\(0.63x = 0.27\)

\(x = \frac{0.27}{0.63} = \frac{3}{7}\)

The total number of sweets is 52 (13 red + 13 blue + 13 green + 13 yellow). We need to find the probability of selecting exactly 3 red sweets out of 7.

Using combinations, the number of ways to choose 3 red sweets from 13 is given by:

\(\binom{13}{3}\)

The number of ways to choose the remaining 4 sweets from the 39 non-red sweets is:

\(\binom{39}{4}\)

The total number of ways to choose 7 sweets from 52 is:

\(\binom{52}{7}\)

Thus, the probability is:

\(\frac{\binom{13}{3} \times \binom{39}{4}}{\binom{52}{7}}\)

Calculating these values:

\(\binom{13}{3} = 286\)

\(\binom{39}{4} = 82,251\)

\(\binom{52}{7} = 133,784,560\)

Therefore, the probability is:

\(\frac{286 \times 82,251}{133,784,560} \approx 0.176\)

Each die has 4 faces, so there are a total of 16 possible outcomes when two dice are thrown together.

To find the probability of the sum being 7 or more, we consider the following outcomes:

• Sum of 7: Possible outcomes are (3,4) and (4,3), giving a probability of \(\frac{2}{16}\).
• Sum of 8: The only possible outcome is (4,4), giving a probability of \(\frac{1}{16}\).

Thus, the probability of the sum being 7 or more is:

\(P(7 \text{ or more}) = \frac{2}{16} + \frac{1}{16} = \frac{3}{16}\)

The expected number of occasions when the sum is 7 or more in 200 throws is:

\(200 \times \frac{3}{16} = 37.5\)

(i) To find the probability that the three peppers are all different colors, we calculate the number of ways to choose one pepper of each color and divide by the total number of ways to choose any three peppers.

The number of ways to choose one red, one green, and one yellow pepper is given by:

\(\binom{3}{1} \times \binom{4}{1} \times \binom{5}{1} = 3 \times 4 \times 5 = 60\)

The total number of ways to choose any three peppers from twelve is:

\(\binom{12}{3} = 220\)

Thus, the probability is:

\(\frac{60}{220} = \frac{3}{11}\)

(ii) To find the probability that exactly 2 of the peppers taken are green, we calculate the number of ways to choose 2 green peppers and 1 non-green pepper, and divide by the total number of ways to choose any three peppers.

The number of ways to choose 2 green peppers is:

\(\binom{4}{2} = 6\)

The number of ways to choose 1 non-green pepper (from the remaining 8 peppers) is:

\(\binom{8}{1} = 8\)

Thus, the number of favorable outcomes is:

\(6 \times 8 = 48\)

The probability is:

\(\frac{48}{220} = \frac{12}{55}\)

(i) The total probability of having burgers is given by:

\(0.8x + (0.2)(0.75) = 0.5\)

Solving for \(x\):

\(0.8x + 0.15 = 0.5\)

\(0.8x = 0.35\)

\(x = \frac{0.35}{0.8} = 0.438\)

(ii) To find the probability that Henk went swimming given that he had burgers, use Bayes' theorem:

\(P(S \mid B) = \frac{P(S \cap B)}{P(B)}\)

\(P(S \cap B) = (0.2)(0.75) = 0.15\)

\(P(S \mid B) = \frac{0.15}{0.5} = 0.3\)

(i) The probability that both sweets are toffees is calculated by considering each box separately and then summing the probabilities:

For Box A: \(\frac{1}{3} \times \frac{6}{10} \times \frac{5}{9}\)

For Box B: \(\frac{1}{3} \times \frac{5}{8} \times \frac{4}{7}\)

For Box C: \(\frac{1}{3} \times \frac{3}{10} \times \frac{2}{9}\)

Summing these probabilities gives:

\(\frac{1}{3} \times \frac{6}{10} \times \frac{5}{9} + \frac{1}{3} \times \frac{5}{8} \times \frac{4}{7} + \frac{1}{3} \times \frac{3}{10} \times \frac{2}{9} = \frac{53}{210} \approx 0.252\)

(ii) Given that both sweets are toffees, the probability that they came from Box A is calculated using conditional probability:

\(P(A \mid \text{TT}) = \frac{P(A \cap \text{TT})}{P(\text{TT})} = \frac{0.111}{0.252} = \frac{70}{159} \approx 0.440\)

(i) List all possible outcomes of drawing three balls: 123, 124, 125, 134, 135, 145, 234, 235, 245, 345. There are 10 possible outcomes.

To find the probability that the sum is odd, count the outcomes where the sum is odd: 124, 134, 145, 234. There are 4 such outcomes.

The probability is given by \(\frac{4}{10} = 0.4\).

(ii) To find the probability that the largest number \(L\) is 4, consider the outcomes where 4 is the largest: 134, 145, 234, 245. There are 3 such outcomes.

The probability is given by \(\frac{3}{10} = 0.3\).

To find the probability of getting exactly one new pen, we consider the two cases: one new pen and one old pen.

The probability of selecting one new pen first and then one old pen is:

\(P(N, \overline{N}) = \frac{3}{10} \times \frac{7}{9} = \frac{21}{90} = \frac{7}{30}\)

The probability of selecting one old pen first and then one new pen is:

\(P(\overline{N}, N) = \frac{7}{10} \times \frac{3}{9} = \frac{21}{90} = \frac{7}{30}\)

Adding these probabilities gives:

\(P(\text{exactly one new pen}) = \frac{7}{30} + \frac{7}{30} = \frac{14}{30} = \frac{7}{15}\)

Alternatively, using combinations:

Total ways to choose 2 pens from 10 is \(\binom{10}{2} = 45\).

Ways to choose 1 new pen from 3 and 1 old pen from 7 is \(\binom{3}{1} \times \binom{7}{1} = 3 \times 7 = 21\).

Thus, the probability is:

\(\frac{21}{45} = \frac{7}{15}\)

(i) To find the probability of obtaining a total score of 5, we list all possible combinations of scores on the three dice that sum to 5:

• (1, 1, 3)
• (1, 2, 2)
• (1, 3, 1)
• (2, 1, 2)
• (2, 2, 1)
• (3, 1, 1)

There are 6 combinations. Since each die has 6 faces, the total number of possible outcomes when throwing three dice is \(6^3 = 216\). Thus, the probability is \(\frac{6}{216} = \frac{1}{36}\).

(ii) To find the probability of obtaining a total score of 7, we list all possible combinations of scores on the three dice that sum to 7:

• (1, 2, 4) and permutations: 6 combinations
• (1, 3, 3) and permutations: 3 combinations
• (2, 2, 3) and permutations: 3 combinations
• (1, 1, 5) and permutations: 3 combinations

There are 15 combinations. Thus, the probability is \(\frac{15}{216} = \frac{5}{72}\).

(a) Let the probability that a student sings in the choir be denoted by P(C). We know that:

\(P(not C) = 0.58\)

\(Therefore, P(C) = 1 - 0.58 = 0.42\)

\(The probability that a student plays in the band and sings in the choir is 0.6 × 0.3 = 0.18.\)

The probability that a student does not play in the band and sings in the choir is 0.4 × x.

Thus, the total probability of singing in the choir is:

\(0.18 + 0.4x = 0.42\)

Solving for x:

\(0.4x = 0.42 - 0.18\)

\(0.4x = 0.24\)

\(x = \frac{0.24}{0.4} = 0.6\)

\((b) The probability that one student plays in the band and sings in the choir is 0.6 × 0.3 = 0.18.\)

The probability that both students play in the band and both sing in the choir is:

\((0.18)^2 = 0.0324\)

The number of students at Canton college who prefer hockey is 56, as shown in the table.

The total number of students surveyed is 500.

The probability that a randomly chosen student is at Canton college and prefers hockey is given by the ratio of the number of Canton hockey students to the total number of students:

\(\text{Probability} = \frac{56}{500}\)

This fraction can be simplified to \(\frac{14}{125}\).

Alternatively, the decimal form of the probability is 0.112.

Let the probability that Shona is late be the complement of the probability that she is not late. Therefore, the probability that she is late is:

\(1 - 0.63 = 0.37\)

The probability that Shona is late if she catches the bus is:

\(0.8 imes 0.4 = 0.32\)

The probability that Shona is late if she cycles is:

0.2 imes x

Thus, the total probability that Shona is late is:

\(0.32 + 0.2x = 0.37\)

Solving for x:

\(0.2x = 0.37 - 0.32\)

\(0.2x = 0.05\)

\(x = rac{0.05}{0.2} = 0.25\)

The probability of choosing a plum from Jameel's box is \(\frac{5}{8}\) because there are 5 plums out of a total of 8 fruits.

The probability of choosing a plum from Rosa's box is \(\frac{x}{x+6}\) because there are x plums out of a total of \(x + 6\) fruits.

The probability that both fruits chosen are plums is given by:

\(\frac{5}{8} \times \frac{x}{x+6} = \frac{1}{4}\)

To solve for x, multiply both sides by \(8(x+6)\):

\(5x = 2(x+6)\)

Expand and simplify:

\(5x = 2x + 12\)

Subtract \(2x\) from both sides:

\(3x = 12\)

Divide both sides by 3:

\(x = 4\)

(i) Total number of selections = \(\binom{12}{7} = 792\).

Selections with the boy included = \(\binom{11}{6} = 462\).

Probability = \(\frac{462}{792} = \frac{7}{12}\).

(ii) Method 1: Scenarios are:

• 2G + 5B: \(\binom{4}{2} \times \binom{8}{5} = 336\)
• 3G + 4B: \(\binom{4}{3} \times \binom{8}{4} = 280\)
• 4G + 3B: \(\binom{4}{4} \times \binom{8}{3} = 56\)

\(Total = 672.\)

Probability = \(\frac{672}{792} = \frac{28}{33}\).

Let \(C\) represent completing the puzzle and \(C'\) represent not completing it.

The probability that Kenny completes the puzzle on both days is \(P(C, C) = 0.8 \times 0.9 = 0.72\).

The probability that Kenny completes the puzzle on Monday but not on Tuesday is \(P(C, C') = 0.8 \times 0.1 = 0.08\).

The probability that Kenny does not complete the puzzle on Monday but completes it on Tuesday is \(P(C', C) = 0.2 \times 0.6 = 0.12\).

The probability that Kenny completes the puzzle on at least one of the two days is:

\(P(C, C) + P(C, C') + P(C', C) = 0.72 + 0.08 + 0.12 = 0.92\)

Alternatively, the probability that Kenny does not complete the puzzle on both days is \(P(C', C') = 0.2 \times 0.4 = 0.08\).

Thus, the probability that he completes the puzzle on at least one day is:

\(1 - P(C', C') = 1 - 0.08 = 0.92\)