Exercise 7.4 Practice
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Overview
This page provides comprehensive Ch 7: The Mathematics of Maybe: Introduction to Probability - Exercise Set 7.4 Practice. Practice constructing branching tree diagrams for multi-stage selection with and without replacement, specifying compound outcomes, and calculating joint probabilities with step-by-step solutions.
Tree Diagrams, Independent Multi-Stage Experiments & Probability Products
Q1: Fruit Basket Tree Diagram
There are two fruit baskets A and B. Basket A has one apple ($A$) and two oranges ($O_1, O_2$). Basket B has one banana ($B$) and one mango ($M$). You randomly pick one fruit from each basket.
(i) Draw a tree diagram showing all possible pairs of fruits.
(ii) List the sample space.
(iii) What is the probability of picking one apple and one banana?
(i) Draw a tree diagram showing all possible pairs of fruits.
(ii) List the sample space.
(iii) What is the probability of picking one apple and one banana?
(i) Tree Diagram:
• First choice (Basket A): apple ($A$), orange 1 ($O_1$), or orange 2 ($O_2$). Each has a probability of $1/3$.
• Second choice (Basket B): banana ($B$) or mango ($M$). Each has a probability of $1/2$.
Branches:
— $A \to B$ (Apple, Banana)
— $A \to M$ (Apple, Mango)
— $O_1 \to B$ (Orange 1, Banana)
— $O_1 \to M$ (Orange 1, Mango)
— $O_2 \to B$ (Orange 2, Banana)
— $O_2 \to M$ (Orange 2, Mango)
• First choice (Basket A): apple ($A$), orange 1 ($O_1$), or orange 2 ($O_2$). Each has a probability of $1/3$.
• Second choice (Basket B): banana ($B$) or mango ($M$). Each has a probability of $1/2$.
Branches:
— $A \to B$ (Apple, Banana)
— $A \to M$ (Apple, Mango)
— $O_1 \to B$ (Orange 1, Banana)
— $O_1 \to M$ (Orange 1, Mango)
— $O_2 \to B$ (Orange 2, Banana)
— $O_2 \to M$ (Orange 2, Mango)
(ii) Sample Space $S$:
$$S = \mathbf{\{(A, B), (A, M), (O_1, B), (O_1, M), (O_2, B), (O_2, M)\}} \implies n(S) = 6$$
$$S = \mathbf{\{(A, B), (A, M), (O_1, B), (O_1, M), (O_2, B), (O_2, M)\}} \implies n(S) = 6$$
(iii) P(Apple and Banana):
Favourable outcome is $\{(A, B)\} \implies n(E) = 1$.
$$P(\text{Apple and Banana}) = \frac{n(E)}{n(S)} = \mathbf{\frac{1}{6} \approx 0.167}$$
Favourable outcome is $\{(A, B)\} \implies n(E) = 1$.
$$P(\text{Apple and Banana}) = \frac{n(E)}{n(S)} = \mathbf{\frac{1}{6} \approx 0.167}$$
Sample space size = 6; P(Apple, Banana) = 1/6 (0.167)
Q2: Double Pen Pick with Replacement
You have a box containing **3 red pens**, **4 black pens**, and **2 green pens**. You pick a pen (without looking) from the box and put it back. Then your friend does the same.
(i) What are the possible outcomes of the pen colours? Can you draw a tree diagram representing the possible outcomes?
(ii) Can you use the tree diagram to guess the probability that both you and your friend pick pens of the same colour?
(i) What are the possible outcomes of the pen colours? Can you draw a tree diagram representing the possible outcomes?
(ii) Can you use the tree diagram to guess the probability that both you and your friend pick pens of the same colour?
Total Pens: $3\text{ Red} + 4\text{ Black} + 2\text{ Green} = 9\text{ pens}$.
Probability for first pick: $P(\text{Red}) = 3/9 = 1/3$, $P(\text{Black}) = 4/9$, $P(\text{Green}) = 2/9$.
Since the pen is replaced, the probabilities remain identical for the second pick.
Probability for first pick: $P(\text{Red}) = 3/9 = 1/3$, $P(\text{Black}) = 4/9$, $P(\text{Green}) = 2/9$.
Since the pen is replaced, the probabilities remain identical for the second pick.
(i) Outcomes / Tree Diagram:
The possible color combinations are:
$$S = \mathbf{\{RR, RB, RG, BR, BB, BG, GR, GB, GG\}}$$
Tree diagram branching:
• First Pick: Red ($1/3$), Black ($4/9$), Green ($2/9$).
• Second Pick: From each of the above, branch out to Red ($1/3$), Black ($4/9$), Green ($2/9$).
The possible color combinations are:
$$S = \mathbf{\{RR, RB, RG, BR, BB, BG, GR, GB, GG\}}$$
Tree diagram branching:
• First Pick: Red ($1/3$), Black ($4/9$), Green ($2/9$).
• Second Pick: From each of the above, branch out to Red ($1/3$), Black ($4/9$), Green ($2/9$).
(ii) Probability of Same Colour:
This happens when outcomes are $RR$, $BB$, or $GG$.
$$P(\text{Same Colour}) = P(RR) + P(BB) + P(GG)$$
$$P(RR) = \frac{3}{9} \times \frac{3}{9} = \frac{9}{81}$$
$$P(BB) = \frac{4}{9} \times \frac{4}{9} = \frac{16}{81}$$
$$P(GG) = \frac{2}{9} \times \frac{2}{9} = \frac{4}{81}$$
$$P(\text{Same Colour}) = \frac{9 + 16 + 4}{81} = \mathbf{\frac{29}{81} \approx 0.358}$$
This happens when outcomes are $RR$, $BB$, or $GG$.
$$P(\text{Same Colour}) = P(RR) + P(BB) + P(GG)$$
$$P(RR) = \frac{3}{9} \times \frac{3}{9} = \frac{9}{81}$$
$$P(BB) = \frac{4}{9} \times \frac{4}{9} = \frac{16}{81}$$
$$P(GG) = \frac{2}{9} \times \frac{2}{9} = \frac{4}{81}$$
$$P(\text{Same Colour}) = \frac{9 + 16 + 4}{81} = \mathbf{\frac{29}{81} \approx 0.358}$$
P(same colour) = 29/81 (approx 0.358)