To find the probability that the two marbles come from bag B given that one is white and one is red, we use conditional probability:

\(P(B \mid WR \text{ or } RW) = \frac{P(W \& R \text{ from bag B})}{P(W \text{ and } R)}\)

First, calculate \(P(W \& R \text{ from bag B})\):

\(P(W \& R \text{ from bag B}) = \frac{2}{3} \times \frac{6}{15} \times \frac{7}{14} + \frac{2}{3} \times \frac{7}{15} \times \frac{6}{14} = \frac{4}{15}\)

Next, calculate \(P(W \text{ and } R)\):

\(P(W \text{ and } R) = P(W \& R \text{ from bag A}) + P(W \& R \text{ from bag B})\)

\(P(W \& R \text{ from bag A}) = \frac{2}{6} \times \frac{8}{15} \times \frac{4}{14} + \frac{2}{6} \times \frac{4}{15} \times \frac{8}{14} = \frac{116}{315}\)

\(P(W \text{ and } R) = \frac{116}{315} + \frac{4}{15} = \frac{232}{630}\)

Finally, calculate the conditional probability:

\(P(B \mid WR \text{ or } RW) = \frac{\frac{4}{15}}{\frac{232}{630}} = \frac{168}{232} = 0.724\)

(a) The probabilities for the tree diagram are:

• From Box A to Box B (Red to Red): \(\frac{x+1}{x+10}\)
• From Box A to Box B (Red to Blue): \(\frac{9}{x+10}\)
• From Box A to Box B (Blue to Red): \(\frac{x}{x+10}\)
• From Box A to Box B (Blue to Blue): \(\frac{10}{x+10}\)

(b) The probability that both balls chosen are blue is:

\(\frac{4}{10} \times \frac{10}{x+10} = \frac{4}{x+10}\)

(c) Given \(\frac{4}{x+10} = \frac{1}{6}\), solve for \(x\):

\(\frac{4}{x+10} = \frac{1}{6}\)

\(4 \times 6 = x + 10\)

\(x + 10 = 24\)

\(x = 14\)

Now, find the probability that the ball chosen from box A is red given that the ball chosen from box B is red:

\(P(A \text{ Red } | B \text{ Red }) = \frac{P(A \text{ Red } \cap B \text{ Red })}{P(B \text{ Red })}\)

\(P(A \text{ Red } \cap B \text{ Red }) = \frac{6}{10} \times \frac{x+1}{x+10} = \frac{6}{10} \times \frac{15}{24} = \frac{3}{8}\)

\(P(B \text{ Red }) = \frac{6}{10} \times \frac{15}{24} + \frac{4}{10} \times \frac{14}{24} = \frac{8}{73}\)

\(P(A \text{ Red } | B \text{ Red }) = \frac{3/8}{8/73} = \frac{45}{73} \approx 0.616\)

To find the probability that a student is male given that they play the drums, we use conditional probability:

\(P(M \mid D) = \frac{P(M \cap D)}{P(D)}\)

From the table, the number of male students who play the drums is 11. The total number of students who play the drums is 11 (male) + 20 (female) = 31.

Thus, \(P(M \cap D) = \frac{11}{180}\) and \(P(D) = \frac{31}{180}\).

Substituting these into the formula gives:

\(P(M \mid D) = \frac{\frac{11}{180}}{\frac{31}{180}} = \frac{11}{31}\)

Let \(C\), \(T\), and \(M\) represent the events of having chocolate, tea, and milk, respectively. Let \(B\) represent the event of having a biscuit.

The probability that Suki does not have a biscuit, \(P(B')\), is calculated as follows:

\(P(B') = P(C \cap B') + P(T \cap B') + P(M \cap B')\)

\(P(C \cap B') = 0.45 \times (1 - 0.3) = 0.45 \times 0.7 = 0.315\)

\(P(T \cap B') = 0.35 \times (1 - 0.6) = 0.35 \times 0.4 = 0.14\)

\(P(M \cap B') = 0.2 \times 1 = 0.2\)

Thus, \(P(B') = 0.315 + 0.14 + 0.2 = 0.655\)

We need to find \(P(T | B')\), the probability that Suki has tea given that she does not have a biscuit:

\(P(T | B') = \frac{P(T \cap B')}{P(B')} = \frac{0.14}{0.655}\)

Calculating this gives:

\(P(T | B') = \frac{0.14}{0.655} \approx 0.214\)

Therefore, the probability that Suki has tea given that she does not have a biscuit is \(0.214\) or \(\frac{28}{131}\).

(i) The number of households in Shan with poor broadband service is 40. The total number of households is 800. Therefore, the probability is given by:

\(P(\text{Shan and Poor}) = \frac{40}{800} = \frac{1}{20}\)

(ii) The number of households in Shan with good broadband service is 177. The total number of households in Shan is \(223 + 177 + 40 = 440\). Therefore, the probability is given by:

\(P(\text{Good} | \text{Shan}) = \frac{177}{440}\)