Let the events be defined as follows:

• ext{SSS}: All three vehicles go straight on.
• ext{LLL}: All three vehicles turn left.
• ext{RRR}: All three vehicles turn right.

The probability that all three vehicles choose the same direction is given by:

\(P( ext{same direction}) = P( ext{SSS}) + P( ext{LLL}) + P( ext{RRR})\)

Calculating each probability:

\(P( ext{SSS}) = 0.3^3\)

\(P( ext{LLL}) = 0.55^3\)

\(P( ext{RRR}) = 0.15^3\)

Thus,

\(P( ext{same direction}) = 0.3^3 + 0.55^3 + 0.15^3\)

We need to find the probability that all three vehicles go straight on given that they all choose the same direction:

\(P( ext{SSS} \,|\, ext{all same direction}) = \frac{P( ext{SSS and same direction})}{P( ext{same direction})}\)

Since ext{SSS} is a subset of ext{same direction},

\(P( ext{SSS and same direction}) = P( ext{SSS}) = 0.3^3\)

Therefore,

\(P( ext{SSS} \,|\, ext{all same direction}) = \frac{0.3^3}{0.3^3 + 0.55^3 + 0.15^3}\)

Calculating the values:

\(0.3^3 = 0.027\)

\(0.55^3 = 0.166375\)

\(0.15^3 = 0.003375\)

Thus,

\(P( ext{same direction}) = 0.027 + 0.166375 + 0.003375 = 0.197\)

Finally,

\(P( ext{SSS} \,|\, ext{all same direction}) = \frac{0.027}{0.197} \approx 0.137\)

(i) The tree diagram is as follows:

First attempt: Pass (0.85), Fail (0.15)

If Fail: Retake Pass (0.65), Retake Fail (0.35)

(ii) We need to find the probability that a student fails the first attempt given that they gain the certificate.

Let \(F\) be the event of failing the first attempt and \(P\) be the event of passing overall.

The probability of passing overall, \(P(P)\), is given by:

\(P(P) = P(\text{Pass first attempt}) + P(\text{Fail first attempt and Pass retake})\)

\(P(P) = 0.85 + (0.15 \times 0.65) = 0.85 + 0.0975 = 0.9475\)

The probability of failing the first attempt and passing the retake, \(P(F \cap P)\), is:

\(P(F \cap P) = 0.15 \times 0.65 = 0.0975\)

Using conditional probability:

\(P(F | P) = \frac{P(F \cap P)}{P(P)} = \frac{0.0975}{0.9475} = \frac{39}{379} \approx 0.103\)

To find \(P(B \,|\, A')\), we use the formula for conditional probability:

\(P(B \,|\, A') = \frac{P(B \cap A')}{P(A')}\)

First, calculate \(P(A')\):

The total number of outcomes when throwing two dice is 36. The event \(A\) (sum less than 6) includes the outcomes: (1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1). There are 10 outcomes, so \(P(A) = \frac{10}{36}\).

Thus, \(P(A') = 1 - P(A) = 1 - \frac{10}{36} = \frac{26}{36}\).

Next, calculate \(P(B \cap A')\):

The event \(B\) (difference at most 2) includes outcomes where the difference between the numbers is 0, 1, or 2. The outcomes that satisfy \(B \cap A'\) are those that are not in \(A\) but satisfy \(B\). There are 16 such outcomes.

Thus, \(P(B \cap A') = \frac{16}{36}\).

Now, substitute these into the conditional probability formula:

\(P(B \,|\, A') = \frac{\frac{16}{36}}{\frac{26}{36}} = \frac{16}{26} = \frac{8}{13}\)

(i) The tree diagram is as follows:

• First branch: Economy (E) with probability 0.82, First class (F) with probability 0.18.
• Second branch from Economy: Good Night's Sleep (GNS) with probability \(x\), Not GNS with probability \(1-x\).
• Second branch from First class: GNS with probability 0.9, Not GNS with probability 0.1.

(ii) The probability of getting a good night's sleep is given by:

\(0.82x + 0.18 imes 0.9 = 0.285\)

Solving for \(x\):

\(0.82x + 0.162 = 0.285\)

\(0.82x = 0.123\)

\(x = \frac{0.123}{0.82} = 0.15\)

(iii) We need to find \(P(E | \text{not GNS})\):

\(P(E | \text{not GNS}) = \frac{P(E \cap \text{not GNS})}{P(\text{not GNS})}\)

\(P(E \cap \text{not GNS}) = 0.82 imes (1 - 0.15) = 0.82 imes 0.85 = 0.697\)

\(P(\text{not GNS}) = 1 - 0.285 = 0.715\)

\(P(E | \text{not GNS}) = \frac{0.697}{0.715} = 0.975\)

(i) To find the probability that the jar is small, we calculate the probability of a jar being small for both Café Premium and Café Standard and sum them:

\(P(S) = 0.65 \times 0.6 + 0.35 \times 0.75\)

\(= 0.39 + 0.2625 = 0.6525\)

Therefore, the probability that the jar is small is approximately 0.653.

(ii) To find the probability that the jar is Café Standard given that it is large, we use conditional probability:

\(P(\text{Std} \mid L) = \frac{P(\text{Std} \cap L)}{P(L)}\)

Where:

\(P(\text{Std} \cap L) = 0.35 \times 0.25 = 0.0875\)

And:

\(P(L) = 1 - P(S) = 1 - 0.6525 = 0.3475\)

Thus:

\(P(\text{Std} \mid L) = \frac{0.0875}{0.3475} \approx 0.252\)

Therefore, the probability that the jar is Café Standard given that it is large is approximately 0.252.