Q28. Box A contains four balls labelled and 5, Box contains six balls labelled
and 8 . A ball is drawn at random from each box.
(a) Draw a possibility diagram to show the outcomes of the draw.
(b) Find, as a fraction in its simplest form, the probability that
(i) both balls have the same number,
(ii) the sum of the two numbers is 10 ,
(iii) the product of the two numbers is at most 15 ,
(iv) at least one of the numbers is a perfect square.

Ask by Valdez Spencer. in Singapore

Mar 29,2025

Question

Q28. Box A contains four balls labelled and 5, Box contains six balls labelled
and 8 . A ball is drawn at random from each box.
(a) Draw a possibility diagram to show the outcomes of the draw.
(b) Find, as a fraction in its simplest form, the probability that
(i) both balls have the same number,
(ii) the sum of the two numbers is 10 ,
(iii) the product of the two numbers is at most 15 ,
(iv) at least one of the numbers is a perfect square.

Ask by Valdez Spencer. in Singapore

Mar 29,2025

UpStudy AI · Accepted Answer

**(a) Possibility Diagram**

The outcomes can be arranged in a table where the rows represent the ball drawn from Box A and the columns represent the ball drawn from Box B.

$$
\begin{array}{c|cccccc}
\text{Box A} \backslash \text{Box B} & 3 & 4 & 5 & 6 & 7 & 8 \ \hline
2 & (2,3) & (2,4) & (2,5) & (2,6) & (2,7) & (2,8) \
3 & (3,3) & (3,4) & (3,5) & (3,6) & (3,7) & (3,8) \
4 & (4,3) & (4,4) & (4,5) & (4,6) & (4,7) & (4,8) \
5 & (5,3) & (5,4) & (5,5) & (5,6) & (5,7) & (5,8) \
\end{array}
$$

There are a total of $$4 \times 6 = 24$$ outcomes.

---

**(b) Probabilities**

**(i) Both balls have the same number**

The numbers common to both boxes are $$3, 4, 5$$. The favorable outcomes are:
$$
(3,3),\ (4,4),\ (5,5)
$$
There are $$3$$ outcomes.

Thus the probability is 
$$
\frac{3}{24} = \frac{1}{8}.
$$

---

**(ii) The sum of the two numbers is 10**

We look for pairs $$(a,b)$$ such that $$a+b=10$$.

Checking each possibility from Box A:
- For $$a=2$$: $$b$$ must be $$8$$ since $$2+8=10$$. Outcome: $$(2,8)$$.
- For $$a=3$$: $$b$$ must be $$7$$ since $$3+7=10$$. Outcome: $$(3,7)$$.
- For $$a=4$$: $$b$$ must be $$6$$ since $$4+6=10$$. Outcome: $$(4,6)$$.
- For $$a=5$$: $$b$$ must be $$5$$ since $$5+5=10$$. Outcome: $$(5,5)$$.

There are $$4$$ favorable outcomes.

Thus the probability is 
$$
\frac{4}{24} = \frac{1}{6}.
$$

---

**(iii) The product of the two numbers is at most 15**

Examine each possibility by ball drawn from Box A:

- For $$a=2$$:
  - $$2 \times 3 = 6 \le 15$$
  - $$2 \times 4 = 8 \le 15$$
  - $$2 \times 5 = 10 \le 15$$
  - $$2 \times 6 = 12 \le 15$$
  - $$2 \times 7 = 14 \le 15$$
  - $$2 \times 8 = 16 > 15$$  
  Favorable outcomes from $$a=2$$: $$5$$.

- For $$a=3$$:
  - $$3 \times 3 = 9 \le 15$$
  - $$3 \times 4 = 12 \le 15$$
  - $$3 \times 5 = 15 \le 15$$
  - $$3 \times 6 = 18 > 15$$
  - $$3 \times 7 = 21 > 15$$
  - $$3 \times 8 = 24 > 15$$  
  Favorable outcomes from $$a=3$$: $$3$$.

- For $$a=4$$:
  - $$4 \times 3 = 12 \le 15$$
  - $$4 \times 4 = 16 > 15$$  
  (Further products exceed 15.)  
  Favorable outcomes from $$a=4$$: $$1$$.

- For $$a=5$$:
  - $$5 \times 3 = 15 \le 15$$
  - $$5 \times 4 = 20 > 15$$  
  (Further products exceed 15.)  
  Favorable outcomes from $$a=5$$: $$1$$.

Total favorable outcomes: $$5 + 3 + 1 + 1 = 10$$.

Thus the probability is 
$$
\frac{10}{24} = \frac{5}{12}.
$$

---

**(iv) At least one of the numbers is a perfect square**

The only perfect square available in the labelled balls is $$4$$ (since $$2, 3, 5, 6, 7, 8$$ are not perfect squares).

It is easier to compute the complement: neither ball is $$4$$.

- For Box A, excluding $$4$$ leaves $$\{2, 3, 5\}$$ $$\Rightarrow 3$$ outcomes.
- For Box B, excluding $$4$$ leaves $$\{3, 5, 6, 7, 8\}$$ $$\Rightarrow 5$$ outcomes.

Total outcomes with no $$4$$ is $$3 \times 5 = 15$$.

Thus, outcomes with at least one $$4$$ is:
$$
24 - 15 = 9.
$$

The probability is 
$$
\frac{9}{24} = \frac{3}{8}.
$$
**(a) Possibility Diagram** The outcomes are arranged in a table where each row represents a ball drawn from Box A and each column represents a ball drawn from Box B. $$ \begin{array}{c|cccccc} \text{Box A} \backslash \text{Box B} & 3 & 4 & 5 & 6 & 7 & 8 \ \hline 2 & (2,3) & (2,4) & (2,5) & (2,6) & (2,7) & (2,8) \ 3 & (3,3) & (3,4) & (3,5) & (3,6) & (3,7) & (3,8) \ 4 & (4,3) & (4,4) & (4,5) & (4,6) & (4,7) & (4,8) \ 5 & (5,3) & (5,4) & (5,5) & (5,6) & (5,7) & (5,8) \ \end{array} $$ There are a total of $$4 \times 6 = 24$$ possible outcomes. --- **(b) Probabilities** **(i) Both balls have the same number** The numbers common to both boxes are $$3, 4, 5$$. The favorable outcomes are: $$ (3,3),\ (4,4),\ (5,5) $$ There are $$3$$ favorable outcomes. Thus, the probability is $$ \frac{3}{24} = \frac{1}{8}. $$ **(ii) The sum of the two numbers is 10** The favorable outcomes where $$a + b = 10$$ are: $$ (2,8),\ (3,7),\ (4,6),\ (5,5) $$ There are $$4$$ favorable outcomes. Thus, the probability is $$ \frac{4}{24} = \frac{1}{6}. $$ **(iii) The product of the two numbers is at most 15** The favorable outcomes where $$a \times b \leq 15$$ are: $$ (2,3),\ (2,4),\ (2,5),\ (2,6),\ (2,7),\ (3,3),\ (3,4),\ (3,5),\ (4,3),\ (5,3) $$ There are $$10$$ favorable outcomes. Thus, the probability is $$ \frac{10}{24} = \frac{5}{12}. $$ **(iv) At least one of the numbers is a perfect square** The only perfect square available is $$4$$. The number of outcomes where neither ball is $$4$$ is: $$ (2,3),\ (2,5),\ (2,6),\ (2,7),\ (2,8),\ (3,3),\ (3,5),\ (3,6),\ (3,7),\ (3,8),\ (5,3),\ (5,5),\ (5,6),\ (5,7),\ (5,8) $$ There are $$15$$ such outcomes. Thus, the number of outcomes with at least one $$4$$ is: $$ 24 - 15 = 9 $$ The probability is $$ \frac{9}{24} = \frac{3}{8}. $$

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