The Mathematics of Maybe: Introduction to Probability — Important Questions

Answers

Correction for question 15 based on expected interpretation for a 2-mark question for each part:
Let's assume the question means picking one fruit from each basket.
Basket A has 1 Apple and 2 Oranges (Total 3 fruits).
Basket B has 1 Banana and 1 Mango (Total 2 fruits).
(i) Tree Diagram:
From Basket A: Apple (1/3), Orange (2/3)
From Basket B: Banana (1/2), Mango (1/2)
Outcomes:
Apple and Banana: P = (1/3) * (1/2) = 1/6
Apple and Mango: P = (1/3) * (1/2) = 1/6
Orange and Banana: P = (2/3) * (1/2) = 2/6 = 1/3
Orange and Mango: P = (2/3) * (1/2) = 2/6 = 1/3
(ii) Sample Space: {(Apple, Banana), (Apple, Mango), (Orange, Banana), (Orange, Mango)}. There are 4 distinct types of outcomes.
(iii) Probability of picking one apple and one banana = 1/6.

(i) A ball is drawn, recorded, returned, and a second ball is drawn.
Let the numbers be 1, 2, 3, 4.
Possible outcomes for the first draw: {1, 2, 3, 4} (4 outcomes)
Possible outcomes for the second draw: {1, 2, 3, 4} (4 outcomes)
Sample space size = 4 * 4 = 16.
Tree diagram would branch out 4 times from the start, and then each of those would branch out 4 times again.

(ii) A ball is drawn, recorded. Without replacing, a second ball is drawn.
Possible outcomes for the first draw: {1, 2, 3, 4} (4 outcomes)
If the first ball drawn was '1', the second draw can be from {2, 3, 4} (3 outcomes).
This applies for each of the first 4 draws.
Sample space size = 4 * 3 = 12.
(Example: (1,2), (1,3), (1,4), (2,1), (2,3), (2,4), etc.)
Tree diagram would branch out 4 times from the start, and then each of those would branch out 3 times.

(i) Probability of drawing a red ball first, then a blue ball.
P(1st Red) = 4/9.
After drawing one red ball, there are 3 red balls and 5 blue balls left (total 8 balls).
P(2nd Blue | 1st Red) = 5/8.
P(1st Red and 2nd Blue) = P(1st Red) * P(2nd Blue | 1st Red) = (4/9) * (5/8) = 20/72 = 5/18.

(ii) Probability of drawing 2 blue balls.
P(1st Blue) = 5/9.
After drawing one blue ball, there are 4 red balls and 4 blue balls left (total 8 balls).
P(2nd Blue | 1st Blue) = 4/8 = 1/2.
P(1st Blue and 2nd Blue) = P(1st Blue) * P(2nd Blue | 1st Blue) = (5/9) * (4/8) = 20/72 = 5/18.

• No heads (TTT)
• One head (HTT, THT, TTH)
• Two heads (HHT, HTH, THH)
• Three heads (HHH)
  The number of heads can be 0, 1, 2, or 3.
  List (iv) {0, 1, 2, 3} is the correct sample space because it lists all possible numbers of heads and each value represents a distinct outcome.
• List (i) {1, 2, 3} fails because it omits the possibility of getting 0 heads.
• List (ii) {0, 1, 2} fails because it omits the possibility of getting 3 heads.
• List (iii) {0, 1, 2, 3, 4} fails because it includes 4, which is impossible when tossing only three coins.

Self-correction: The diagram shows a rectangle with one side labelled '2 m'. The other side is not explicitly labelled but appears to be the same length as the diameter of the circle inside, which is 1 m. So, let's assume the rectangle is 2 m by 1 m.
Area of the rectangular region = Length × Width = 2 m × 1 m = 2 m².
The circle has a diameter of 1 m, so its radius is 0.5 m.
Area of the circle = $\pi \ast r^2=\pi \ast (0.5m{)}^2=\pi \ast 0.25m^2=0.25\pi m^2$.
The probability that the die will land inside the circle is:
P(inside circle) = Area of circle / Area of rectangle
P(inside circle) = $(0.25\pi m^2)\mathrm{/}(2m^2)$
P(inside circle) = $0.25\pi \mathrm{/}2=\pi \mathrm{/}8$.
Using $\pi \approx 3.14$:
P(inside circle) $\approx 3.14\mathrm{/}8\approx 0.3925$.