Chapter 13 · Probability

Exercise 13.2 · Q13: Two balls drawn with replacement: red and black

Exercise 13.2Independent events · Section 13.4

The question

Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that

(i) both balls are red.

(ii) first ball is black and second is red.

(iii) one of them is black and other is red.

Exam answer

Let $B_1$ ⁠, $R_1$ be the events that the first ball is black, red, and $B_2$ ⁠, $R_2$ the same for the second ball.

The box has 18 balls and the first ball is replaced, so the two draws are independent and

P(B_1) = P(B_2) = \frac{10}{18} = \frac{5}{9}

P(R_1) = P(R_2) = \frac{8}{18} = \frac{4}{9}

(i)

\begin{aligned} P(R_1\cap R_2) &= P(R_1)\,P(R_2) \\ &= \frac{4}{9}\times\frac{4}{9} \end{aligned}

P(both red) $= \frac{16}{81}$ ⁠.

(ii)

\begin{aligned} P(B_1\cap R_2) &= P(B_1)\,P(R_2) \\ &= \frac{5}{9}\times\frac{4}{9} \end{aligned}

Required probability $= \frac{20}{81}$ ⁠.

(iii) One black and one red means $B_1\cap R_2$ or $R_1\cap B_2$ ⁠, which are mutually exclusive.

\begin{aligned} &P(B_1\cap R_2) + P(R_1\cap B_2) \\ &= \frac{5}{9}\times\frac{4}{9} + \frac{4}{9}\times\frac{5}{9} \\ &= \frac{20}{81} + \frac{20}{81} \end{aligned}

$\therefore$ Required probability $= \dfrac{40}{81}$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Because the first ball is put back, the box is exactly the same for the second draw, so the two draws are independent and probabilities simply multiply. Part (iii) has two orders, black then red or red then black; these cannot both happen, so their probabilities add.

Given, to find, and the plan

Given: A box of $10 + 8 = 18$ balls: 10 black, 8 red. Two balls are drawn one after the other, the first put back before the second is drawn.
To find: (i) P(both red), (ii) P(first black and second red), (iii) P(one black and one red, in either order).
Plan: (1) Name the events and find the probability of each colour on one draw. (2) Use independence to multiply. (3) In (iii), add the two orders. (4) Check by counting the $18\times 18 = 324$ ordered draws.

Basics you need

1Classical probability

$P(E) = \frac{n(E)}{n(S)}$ for equally likely outcomes: each of the 18 balls is equally likely to be drawn. Basics: P(E) = n(E)/n(S)

2With replacement means independent

Putting the ball back restores the box, so the second draw is an independent experiment: $P(E\cap F) = P(E)\,P(F)$ (Remark (iii), Section 13.4). Lesson: independent events

3Adding mutually exclusive cases

If two events cannot happen together, the probability that one or the other happens is the sum of their probabilities. Basics: the addition rule

4Tree diagrams

Branch at each draw, multiply along a path, add the paths you want. Basics: tree diagrams

Part (i): both balls are red

Name the events and find the single-draw probabilities

Let $B_1$ and $R_1$ be the events "the first ball is black" and "the first ball is red", and $B_2$ ⁠, $R_2$ the same for the second ball. On any draw all 18 balls are equally likely, so
$\begin{aligned} P(B_1) = P(B_2) &= \frac{10}{18} = \frac{5}{9}, \\ P(R_1) = P(R_2) &= \frac{8}{18} = \frac{4}{9}. \end{aligned}$
The second draw has the same probabilities as the first because the first ball went back into the box: still 10 black and 8 red out of 18.

$P(E) = \frac{n(E)}{n(S)}$ ⁠. Basics: classical probability
Multiply, because the draws are independent
$\begin{aligned} P(R_1\cap R_2) &= P(R_1)\,P(R_2) \\ &= \frac{4}{9}\times\frac{4}{9} = \frac{16}{81}. \end{aligned}$
With replacement, the result of the first draw does not change the box, so $P(R_2|R_1) = P(R_2)$ ⁠: the draws are independent, and the multiplication rule $P(R_1\cap R_2) = P(R_1)\,P(R_2|R_1)$ becomes a plain product.

Definition 3 and Remark (iii), Section 13.4. Lesson: independent events

Part (ii): first black, second red

Multiply along the path black, then red
$\begin{aligned} P(B_1\cap R_2) &= P(B_1)\,P(R_2) \\ &= \frac{5}{9}\times\frac{4}{9} = \frac{20}{81}. \end{aligned}$
The order is fixed here: black must come first. So this is one path of the tree, and the same independence as in Part (i) lets us multiply the two probabilities from Part (i), Step 1.

Multiply along a path. Basics: tree diagrams

Part (iii): one black and one red

Split into the two orders

"One of them is black and the other is red" happens in two ways: black then red ( $B_1\cap R_2$ ⁠), or red then black ( $R_1\cap B_2$ ⁠). The first ball cannot be both black and red, so these two events are mutually exclusive, and
$\begin{aligned} &P(\text{one black, one red}) \\ &\quad = P(B_1\cap R_2) + P(R_1\cap B_2). \end{aligned}$
Unlike Part (ii), this part does not say which ball comes first, so both orders count. Listing the cases that make up the event, and checking they do not overlap, is what allows the addition.

Probabilities of mutually exclusive events add. Basics: the addition rule
Work out each order and add
$\begin{aligned} P(R_1\cap B_2) &= P(R_1)\,P(B_2) \\ &= \frac{4}{9}\times\frac{5}{9} = \frac{20}{81}, \end{aligned}$
and $P(B_1\cap R_2) = \frac{20}{81}$ from Part (ii). So
$\begin{aligned} P(\text{one black, one red}) &= \frac{20}{81} + \frac{20}{81} \\ &= \frac{40}{81}. \end{aligned}$

The tree of the two draws (B = black, R = red). With replacement, every first branch is followed by the same $\frac59$ and $\frac49$ ⁠. The two highlighted paths are the orders of Part (iii): $\frac{20}{81} + \frac{20}{81} = \frac{40}{81}$ ⁠. The four paths add to $\frac{25 + 20 + 20 + 16}{81} = 1$ ⁠.

Each order is one path of the tree, found by multiplying; the event of Part (iii) is two paths, so their probabilities are added.

Multiply along a path, add paths. Basics: tree diagrams

Result

(i) $\frac{16}{81}$ (ii) $\frac{20}{81}$ (iii) $\frac{40}{81}$

(Textbook answer: (i) $\frac{16}{81}$ ⁠, (ii) $\frac{20}{81}$ ⁠, (iii) $\frac{40}{81}$ ⁠.)

Check it with numbers

Count ordered draws: $18\times 18 = 324$ in all. Both red: $8\times 8 = 64$ ⁠, and $\frac{64}{324} \approx 0.1975$ ⁠, the same as $\frac{16}{81} \approx 0.1975$ ⁠. Black then red: $10\times 8 = 80$ ⁠, and $\frac{80}{324} \approx 0.2469$ ⁠. One of each: $80 + 80 = 160$ ⁠, and $\frac{160}{324} \approx 0.4938$ ⁠, the same as $\frac{40}{81} \approx 0.4938$ ⁠.

Common mistakes

Using 17 balls for the second draw. That is drawing without replacement. Here the ball goes back, so it is 18 both times.
Giving $\frac{20}{81}$ for (iii). That counts only the order black then red; red then black counts too.
Multiplying the two orders in (iii). "Black then red" or "red then black" is an "or" of two exclusive events, so the probabilities add.

What to take away

With replacement, draws are independent: multiply the single-draw probabilities.
When the order is not stated, list every order and add the probabilities of these exclusive cases.
A tree diagram keeps both rules in view: multiply along a path, add across paths.

Every fact used, and where it comes from

Fact	Used in	Source
$P(E) = \frac{n(E)}{n(S)}$ for equally likely outcomes	Part (i), Step 1; Check	Class XI, Probability (recalled in Section 13.1)
Independent experiments: $P(E\cap F) = P(E)\,P(F)$	Parts (i), (ii), (iii)	Remark (iii), Section 13.4 (Lesson: independent events)
Multiplication rule $P(E\cap F) = P(E)\,P(F\|E)$	Part (i), Step 2	Section 13.3 (Lesson: the multiplication rule)
Mutually exclusive events: probabilities add	Part (iii)	Class XI, Probability

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Exercise 13.2 · Q13: Two balls drawn with replacement: red and black

Basics you need

Part (i): both balls are red

Name the events and find the single-draw probabilities

Multiply, because the draws are independent

Part (ii): first black, second red

Multiply along the path black, then red

Part (iii): one black and one red

Split into the two orders

Work out each order and add

Every fact used, and where it comes from