Past Exam Questions

To find the probability that the coin shows a head given that Jodie’s score is 8, we use conditional probability:

\(P(H | 8) = \frac{P(H \cap 8)}{P(8)}\)

First, calculate \(P(8)\):

\(P(8) = P(H \text{ and } 4+4) + P(T \text{ and } 2 \times 4) + P(T \text{ and } 4 \times 2)\)

\(= \frac{1}{3} \times \frac{1}{16} + \frac{2}{3} \times \frac{1}{16} + \frac{2}{3} \times \frac{1}{16}\)

\(= \frac{1}{48} + \frac{2}{48} + \frac{2}{48}\)

\(= \frac{5}{48}\)

Next, calculate \(P(H \cap 8)\):

\(P(H \cap 8) = P(H \text{ and } 4+4) = \frac{1}{3} \times \frac{1}{16} = \frac{1}{48}\)

Now, substitute these into the conditional probability formula:

\(P(H | 8) = \frac{\frac{1}{48}}{\frac{5}{48}} = \frac{1}{5}\)

(i) The probability that Ben becomes the champion after playing exactly 2 games is when Ben wins both games. The probability of Ben winning a game is 0.7. Therefore, the probability is:

\(P(B \text{ champ}) = 0.7 \times 0.7 = 0.49\)

(ii) The probability that Ben becomes the champion can occur in the following sequences: WW, WLW, LWW.

\(P(B \text{ champ}) = P(WW) + P(WLW) + P(LWW)\)

\(= (0.7 \times 0.7) + (0.7 \times 0.3 \times 0.7) + (0.3 \times 0.7 \times 0.7)\)

\(= 0.49 + 0.147 + 0.147\)

\(= 0.784\)

(iii) Given that Tom becomes the champion, we need to find the probability that he won the 2nd game. This can occur in the sequences: TT or TBT.

\(P(T2 \cap T) = (0.3 \times 0.3) + (0.7 \times 0.3 \times 0.3) = 0.216\)

The probability that Tom becomes the champion is \(1 - 0.784 = 0.216\).

Thus, the probability is:

\(P(T2 | T) = \frac{P(T2 \cap T)}{P(T)} = \frac{0.216}{0.216} = 0.708\)

To find the probability that Nur chose playground Y given that she chose a climbing frame, we use Bayes' theorem:

\(P(Y | C) = \frac{P(Y \cap C)}{P(C)}\)

First, calculate \(P(Y \cap C)\):

Nur chooses playground Y with probability \(\frac{1}{4}\) and then chooses a climbing frame with probability \(\frac{1}{12}\) (since there is 1 climbing frame out of 12 total pieces of equipment in playground Y).

\(P(Y \cap C) = \frac{1}{4} \times \frac{1}{12} = \frac{1}{48}\)

Next, calculate \(P(C)\), the total probability of choosing a climbing frame:

\(P(C) = P(X \cap C) + P(Y \cap C) + P(Z \cap C)\)

\(P(X \cap C) = 0\) since there are no climbing frames in playground X.

\(P(Y \cap C) = \frac{1}{48}\) as calculated above.

\(P(Z \cap C) = \frac{1}{2} \times \frac{4}{16} = \frac{1}{8}\) since there are 4 climbing frames out of 16 total pieces of equipment in playground Z.

\(P(C) = 0 + \frac{1}{48} + \frac{1}{8} = \frac{1}{48} + \frac{6}{48} = \frac{7}{48}\)

Finally, substitute back into Bayes' theorem:

\(P(Y | C) = \frac{\frac{1}{48}}{\frac{7}{48}} = \frac{1}{7}\)

Let the event that the first digit is 5 be denoted as \(5_1\) and the event that the second digit is not 5 be denoted as \(\overline{5}_2\).

The probability that the first digit is 5 and the second digit is not 5 is:

\(P(5_1 \cap \overline{5}_2) = \frac{3}{9} \times \frac{6}{8} = \frac{1}{4}\)

The probability that the second digit is not 5 is:

\(P(\overline{5}_2) = \frac{1}{4} + \frac{6}{9} \times \frac{5}{8} = \frac{48}{72} = 0.6666\)

The conditional probability that the first digit is 5 given that the second digit is not 5 is:

\(P(5_1 | \overline{5}_2) = \frac{\frac{1}{4}}{\frac{48}{72}} = \frac{3}{8} = 0.375\)

(i) The probability that Mohit goes to the cinema is 0.35. The probability that he spends less than $50 at the cinema is 1 - 0.8 = 0.2. Therefore, the probability that Mohit will go to the cinema and spend less than $50 is:

\(P(C \cap <50) = 0.35 \times 0.2 = 0.07\)

(ii) We need to find \(P(C \mid <50)\), the probability that Mohit went to the cinema given that he spent less than $50. This is given by:

\(P(C \mid <50) = \frac{P(C \cap <50)}{P(<50)}\)

We already found \(P(C \cap <50) = 0.07\).

Now, calculate \(P(<50)\):

\(P(<50) = P(S \cap <50) + P(C \cap <50) + P(H \cap <50)\)

\(P(S \cap <50) = 0.25 \times (1 - 0.7) = 0.25 \times 0.3 = 0.075\)

\(P(C \cap <50) = 0.07\)

\(P(H \cap <50) = 0.4 \times 1 = 0.4\)

Thus, \(P(<50) = 0.075 + 0.07 + 0.4 = 0.545\)

Therefore, \(P(C \mid <50) = \frac{0.07}{0.545} = 0.128\) (or \(\frac{14}{109}\))

Let \(M\) be the event that Mumbok is chosen, and \(18{-}60\) be the event that a person aged between 18 and 60 is chosen.

The probability of choosing Mumbok and a person aged 18 to 60 is:

\(P(M \cap 18{-}60) = 0.6 \times \frac{55}{90} = 0.367 \text{ or } \frac{11}{30}\)

The probability of choosing a person aged 18 to 60 from either town is:

\(P(18{-}60) = 0.6 \times \frac{55}{90} + 0.4 \times \frac{95}{160} = \frac{29}{48} \text{ or } 0.604\)

Using Bayes' theorem, the probability that the town chosen was Mumbok given that the person is aged 18 to 60 is:

\(P(M \mid 18{-}60) = \frac{P(M \cap 18{-}60)}{P(18{-}60)} = \frac{\frac{11}{30}}{\frac{29}{48}} = \frac{88}{145} \text{ or } 0.607\)

Let \(W_1\) be the event that John wins his first game and \(W_2\) be the event that he wins his second game.

The probability that John wins his second game, \(P(W_2)\), can be calculated as:

\(P(W_2) = P(W_1 \cap W_2) + P(L_1 \cap W_2)\)

\(= 0.3 \times 0.6 + 0.7 \times 0.15\)

\(= 0.18 + 0.105\)

\(= 0.285\)

We need to find \(P(W_1 | W_2)\), the probability that John won his first game given that he won his second game. This is given by:

\(P(W_1 | W_2) = \frac{P(W_1 \cap W_2)}{P(W_2)}\)

\(= \frac{0.18}{0.285}\)

\(= 0.632\)

Thus, the probability that John won his first game given that he won his second game is \(0.632\) or \(\frac{12}{19}\).

(a) The probability that Tom removes a red disc on his first turn can be calculated by considering two scenarios: Sam removes a red disc first, then Tom removes a red disc, or Sam removes a white disc first, then Tom removes a red disc.

Probability of Sam removing a red disc first: \(\frac{3}{8}\)

Probability of Tom removing a red disc after Sam: \(\frac{2}{7}\)

Probability of Sam removing a white disc first: \(\frac{5}{8}\)

Probability of Tom removing a red disc after Sam: \(\frac{3}{7}\)

Total probability: \(\frac{3}{8} \times \frac{2}{7} + \frac{5}{8} \times \frac{3}{7} = \frac{21}{56} = \frac{3}{8}\)

(b) Tom wins on his second turn if the sequence of draws is either RRWR, WRRR, or WRWR.

Probability of RRWR: \(\frac{3}{8} \times \frac{2}{7} \times \frac{5}{6} \times \frac{1}{5} = \frac{1}{56}\)

Probability of WRRR: \(\frac{5}{8} \times \frac{3}{7} \times \frac{2}{6} \times \frac{1}{5} = \frac{1}{56}\)

Probability of WRWR: \(\frac{5}{8} \times \frac{3}{7} \times \frac{2}{6} \times \frac{3}{5} = \frac{1}{14}\)

Total probability: \(\frac{1}{56} + \frac{1}{56} + \frac{1}{14} = \frac{3}{28}\)

(c) Given that Tom wins on his second turn, we need to find the probability that Sam removes a red disc on his first turn.

Probability of Sam removing a red disc first: \(\frac{3}{8}\)

Probability of Tom winning on his second turn given Sam removes a red disc first: \(\frac{1}{56}\)

Total probability of Tom winning on his second turn: \(\frac{3}{28}\)

Conditional probability: \(\frac{\frac{1}{56}}{\frac{3}{28}} = \frac{1}{6}\)

(i) The probability of taking a toffee from Susan's bag is \(\frac{5}{12}\). After placing it in Ahmad's bag, the probability of taking a boiled sweet from Ahmad's bag is \(\frac{2}{10}\). Therefore, the probability that the two sweets taken are a toffee from Susan’s bag and a boiled sweet from Ahmad’s bag is:

\(P(T, B) = \frac{5}{12} \times \frac{2}{10} = \frac{1}{12}\)

(ii) Let \(C_S\) be the event that a chocolate is taken from Susan's bag, and \(C_A\) be the event that a chocolate is taken from Ahmad's bag. We need to find \(P(C_S | C_A)\).

First, calculate \(P(C_S \cap C_A)\):

\(P(C_S \cap C_A) = \frac{7}{12} \times \frac{4}{10} = \frac{28}{120}\)

Next, calculate \(P(C_A)\):

\(P(C_A) = \frac{7}{12} \times \frac{4}{10} + \frac{5}{12} \times \frac{3}{10} = \frac{28}{120} + \frac{15}{120} = \frac{43}{120}\)

Finally, calculate \(P(C_S | C_A)\):

\(P(C_S | C_A) = \frac{P(C_S \cap C_A)}{P(C_A)} = \frac{28/120}{43/120} = \frac{28}{43}\)

(i) The total number of balls in box B after transferring one ball from box A is \(5 + x + 1 = x + 6\). Therefore, the probability of choosing a yellow ball from box B is \(\frac{x}{x+6}\).

(ii) The tree diagram is completed as follows:
From box A to box B:
- White to White: \(\frac{6}{x+6}\)
- White to Yellow: \(\frac{x}{x+6}\)
- Yellow to White: \(\frac{5}{x+6}\)
- Yellow to Yellow: \(\frac{x+1}{x+6}\).

(iii) Given that the probability of choosing a white ball from box B after a white ball from box A is \(\frac{1}{3}\), we have:
\(\frac{6}{x+6} = \frac{1}{3}\)
Solving for \(x\):
\(6 = \frac{x+6}{3}\)
\(18 = x + 6\)
\(x = 12\).

(iv) To find the conditional probability that the ball chosen from box A was yellow given that the ball chosen from box B is yellow, we use:
\(P(A_Y | B_Y) = \frac{P(A_Y \cap B_Y)}{P(B_Y)}\)
\(P(B_Y) = \frac{8}{10} \times \frac{12}{18} + \frac{2}{10} \times \frac{13}{18} = \frac{61}{90}\)
\(P(A_Y \cap B_Y) = \frac{2}{10} \times \frac{13}{18} = \frac{13}{90}\)
\(P(A_Y | B_Y) = \frac{13/90}{61/90} = \frac{13}{61}\).

(i) The probability that Fabio chooses Americano and leaves it to drink later is given by the product of the probability of choosing Americano and the probability of leaving it to drink later. This is calculated as:

\(P(A \text{ Later}) = 0.5 \times 0.2 = 0.1\)

(ii) To find the probability that Fabio chose Latte given that he drinks his coffee immediately, we use Bayes' theorem. Let \(L\) be the event that he chooses Latte, and \(I\) be the event that he drinks immediately. We need to find \(P(L \mid I)\).

First, calculate \(P(I)\), the total probability of drinking immediately:

\(P(I) = (0.5 \times 0.8) + (0.3 \times 0.6) + (0.2 \times 0.1) = 0.4 + 0.18 + 0.02 = 0.6\)

Now, calculate \(P(L \cap I)\), the probability that he chooses Latte and drinks immediately:

\(P(L \cap I) = 0.2 \times 0.1 = 0.02\)

Finally, apply Bayes' theorem:

\(P(L \mid I) = \frac{P(L \cap I)}{P(I)} = \frac{0.02}{0.6} = 0.0333\)

Let \(E\), \(L\), and \(OT\) represent the events that Ana is early, late, and on time, respectively. Let \(B\) represent the event that Ana eats a banana, and \(\overline{B}\) represent the event that she does not eat a banana.

We need to find \(P(OT | \overline{B})\).

Using the law of total probability, we have:

\(P(\overline{B}) = P(\overline{B} | E)P(E) + P(\overline{B} | L)P(L) + P(\overline{B} | OT)P(OT)\)

Given:

• \(P(E) = 0.05\)
• \(P(L) = 0.75\)
• \(P(OT) = 1 - P(E) - P(L) = 0.2\)
• \(P(\overline{B} | E) = 1 - 0.7 = 0.3\)
• \(P(\overline{B} | L) = 1\)
• \(P(\overline{B} | OT) = 1 - 0.4 = 0.6\)

Substitute these values into the equation:

\(P(\overline{B}) = 0.3 \times 0.05 + 1 \times 0.75 + 0.6 \times 0.2\)

\(P(\overline{B}) = 0.015 + 0.75 + 0.12 = 0.885\)

Now, use Bayes' theorem to find \(P(OT | \overline{B})\):

\(P(OT | \overline{B}) = \frac{P(\overline{B} | OT)P(OT)}{P(\overline{B})} = \frac{0.6 \times 0.2}{0.885}\)

\(P(OT | \overline{B}) = \frac{0.12}{0.885} \approx 0.136\)

Thus, the probability that Ana is on time given that she does not eat a banana is \(0.136 \left( \frac{8}{59} \right)\).

(i) The probability that the radio is set to station 3 and the presenter is male is given by:

\(0.25p = 0.075\)

Solving for \(p\):

\(p = \frac{0.075}{0.25} = 0.3\)

(ii) To find the probability that the radio was set to station 2 given that Maria hears a male presenter, we use conditional probability:

\(P(2|M) = \frac{P(2 \text{ and } M)}{P(M)}\)

Where:

\(P(2 \text{ and } M) = 0.45 \times 0.85 = 0.3825\)

\(P(M) = 0.3 \times 0.1 + 0.45 \times 0.85 + 0.25 \times 0.3 = 0.4875\)

Thus,

\(P(2|M) = \frac{0.3825}{0.4875} = 0.785\)

Let the event that the pen is in the pencil case be denoted as \(P(C) = 0.7\), and the event that Ted finds the pen given it is in the pencil case be \(P(F|C) = 1\). The event that the pen is not in the pencil case is \(P(C') = 0.3\), and the probability that Ted finds the pen given it is not in the pencil case is \(P(F|C') = 0.2\).

We need to find \(P(C|F)\), the probability that the pen is in the pencil case given that Ted finds it. Using Bayes' theorem:

\(P(C|F) = \frac{P(C \cap F)}{P(F)}\)

Where:

\(P(C \cap F) = P(C) \cdot P(F|C) = 0.7 \times 1 = 0.7\)

\(P(F) = P(C \cap F) + P(C' \cap F) = 0.7 + (0.3 \times 0.2) = 0.7 + 0.06 = 0.76\)

Thus:

\(P(C|F) = \frac{0.7}{0.76} = 0.921\)

(i) The tree diagram is structured as follows:

• First branch: Probability of using a cell phone (Ph) = 0.68, not using a cell phone (NPh) = 0.32.
• Second branch for Ph: Young (Y) = 0.7, Middle-aged (M) = 0.25, Old (O) = 0.05.
• Second branch for NPh: Young (Y) = 0.26, Middle-aged (M) = 0.10, Old (O) = 0.64.

(ii) To find the probability that a passenger aged 45 (middle-aged) uses a cell phone, use conditional probability:

\(P(\text{Ph} \mid \text{M}) = \frac{P(\text{Ph} \cap \text{M})}{P(\text{M})}\)

Where:

• \(P(\text{Ph} \cap \text{M}) = 0.68 \times 0.25 = 0.17\)
• \(P(\text{M}) = (0.68 \times 0.25) + (0.32 \times 0.10) = 0.17 + 0.032 = 0.202\)

Thus,

\(P(\text{Ph} \mid \text{M}) = \frac{0.17}{0.202} \approx 0.842\)

To find the probability that the spinner landed on an even-numbered side given that Raj's score is 12, we use conditional probability:

Let \(E\) be the event that the spinner lands on an even-numbered side. We need to find \(P(E \mid 12)\).

First, calculate \(P(E \text{ and } 12)\):

The probability of landing on an even-numbered side is \(\frac{2}{5}\) (since there are 2 even numbers: 2 and 4).

The possible dice outcomes that multiply to 12 are: (2,6), (6,2), (3,4), (4,3). Each pair has a probability of \(\frac{1}{36}\) (since there are 36 possible outcomes when rolling two dice).

Thus, \(P(E \text{ and } 12) = \frac{2}{5} \times \frac{4}{36} = \frac{8}{180} = \frac{2}{45}\).

Next, calculate \(P(12)\):

For an odd-numbered side, the possible dice outcomes that add to 12 are: (6,6). The probability of this is \(\frac{1}{36}\).

The probability of landing on an odd-numbered side is \(\frac{3}{5}\) (since there are 3 odd numbers: 1, 3, 5).

Thus, \(P(12) = \frac{3}{5} \times \frac{1}{36} + \frac{8}{180} = \frac{11}{180}\).

Finally, use the conditional probability formula:

\(P(E \mid 12) = \frac{P(E \text{ and } 12)}{P(12)} = \frac{\frac{8}{180}}{\frac{11}{180}} = \frac{8}{11}\).

Let \(D\) be the event that the dog chases ducks, and \(G\) be the event that the dog chases geese. Let \(A\) be the event that the dog is attacked, and \(NA\) be the event that the dog is not attacked.

We need to find \(P(G \mid NA)\), which is given by:

\(P(G \mid NA) = \frac{P(G \cap NA)}{P(NA)}\)

First, calculate \(P(G \cap NA)\):

\(P(G \cap NA) = P(G) \times P(NA \mid G) = \frac{2}{5} \times \left(1 - \frac{3}{4}\right) = \frac{2}{5} \times \frac{1}{4} = \frac{1}{10}\)

Next, calculate \(P(NA)\):

\(P(NA) = P(D \cap NA) + P(G \cap NA)\)

\(P(D \cap NA) = P(D) \times P(NA \mid D) = \frac{3}{5} \times \left(1 - \frac{1}{10}\right) = \frac{3}{5} \times \frac{9}{10} = \frac{27}{50}\)

Thus,

\(P(NA) = \frac{27}{50} + \frac{1}{10} = \frac{27}{50} + \frac{5}{50} = \frac{32}{50}\)

Finally, substitute back to find \(P(G \mid NA)\):

\(P(G \mid NA) = \frac{\frac{1}{10}}{\frac{32}{50}} = \frac{1}{10} \times \frac{50}{32} = \frac{5}{32}\)

(i) The probability that the first question is answered correctly is calculated as follows:

\(P(\text{1st correct}) = P(\text{Peter correct}) + P(\text{Peter asks for help and audience correct})\)

\(= 0.7 + 0.2 \times 0.95 = 0.89\)

(ii) To find the probability that the first two questions are both answered correctly, consider the following scenarios:

• Peter answers both questions correctly: \(P(CC) = 0.7 \times 0.7 = 0.49\)
• Peter answers the first correctly, asks for help, and the audience answers correctly: \(P(CHA) = 0.7 \times 0.2 \times 0.95 = 0.133\)
• Peter asks for help on the first, audience answers correctly, and Peter answers the second correctly: \(P(HAC) = 0.2 \times 0.95 \times 0.7 = 0.133\)
• Peter asks for help on the first, audience answers correctly, Peter asks for help on the second, and friend answers correctly: \(P(HAHP) = 0.2 \times 0.95 \times 0.2 \times 0.65 = 0.0247\)

Summing these probabilities gives:

\(P(\text{both correctly answered}) = 0.49 + 0.133 + 0.133 + 0.0247 = 0.781\)

(iii) Given that the first two questions were both answered correctly, the probability that Peter asked the audience is:

\(P(\text{audience} \mid \text{both correct}) = \frac{P(CHA) + P(HAC) + P(HAHP)}{P(\text{both correctly answered})}\)

\(= \frac{0.133 + 0.133 + 0.0247}{0.781} = \frac{0.2907}{0.781} = 0.372\)

(a) The total probability that Kino is late is given by:

\(0.2x + 0.1 \times 2x + 0.7 \times 0.25 = 0.235\)

Simplifying, we have:

\(0.2x + 0.2x + 0.175 = 0.235\)

\(0.4x + 0.175 = 0.235\)

Subtract 0.175 from both sides:

\(0.4x = 0.06\)

Divide by 0.4:

\(x = 0.15\)

(b) We need to find \(P(\text{car} | \text{not late})\).

\(P(\text{car} | \text{not late}) = \frac{P(\text{car and not late})}{P(\text{not late})}\)

\(P(\text{car and not late}) = 0.1 \times (1 - 2x) = 0.1 \times (1 - 0.3) = 0.1 \times 0.7 = 0.07\)

\(P(\text{not late}) = 1 - 0.235 = 0.765\)

Thus,

\(P(\text{car} | \text{not late}) = \frac{0.07}{0.765} = 0.0915\)

Expressed as a fraction, \(\frac{0.07}{0.765} = \frac{70}{765} = \frac{14}{153}\).

(i) The tree diagram is structured as follows:

• First branch: Toast (T) with probability 0.85, Bread roll (B) with probability 0.15.
• Second branch from Toast: Jam (J) with probability 0.8, No Jam (NJ) with probability 0.2.
• Second branch from Bread roll: Jam (J) with probability 0.4, No Jam (NJ) with probability 0.6.

(ii) To find the probability that Maria had toast given she did not have jam, use Bayes' theorem:

\(P(T \mid NJ) = \frac{P(T \text{ and } NJ)}{P(NJ)}\)

Calculate \(P(T \text{ and } NJ)\):

\(P(T \text{ and } NJ) = 0.85 \times 0.2 = 0.17\)

Calculate \(P(NJ)\):

\(P(NJ) = P(T \text{ and } NJ) + P(B \text{ and } NJ) = 0.85 \times 0.2 + 0.15 \times 0.6 = 0.17 + 0.09 = 0.26\)

Substitute back into Bayes' theorem:

\(P(T \mid NJ) = \frac{0.17}{0.26} = \frac{17}{26} \approx 0.654\)