Chapter 13 · Probability

Exercise 13.2 · Q3: Box of oranges approved: three good ones drawn

Exercise 13.2Multiplication rule for three events · Section 13.3

The question

A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.

Exam answer

Let $E$ ⁠, $F$ and $G$ be the events that the first, second and third orange drawn are good.

P(E) = \frac{12}{15}

P(F|E) = \frac{11}{14}

P(G|EF) = \frac{10}{13}

The box is approved only if all three oranges are good. By the multiplication rule of probability,

\begin{aligned} &P(E\cap F\cap G) \\ &= P(E)\,P(F|E)\,P(G|EF) \\ &= \frac{12}{15}\times\frac{11}{14}\times\frac{10}{13} \\ &= \frac{1320}{2730} \end{aligned}

$\therefore$ P(box approved) $= \dfrac{44}{91}$ ⁠.

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

The one idea this problem is built on

"Approved" means good, and good, and good, drawn one after another without putting any back. The multiplication rule for three events multiplies the probability of each draw, given everything drawn before it: $P(E\cap F\cap G) = P(E)\,P(F|E)\,P(G|EF)$ ⁠. Each time a good orange is taken out, the box has one fewer orange and one fewer good one.

Given, to find, and the plan

Given: A box of 15 oranges: 12 good, 3 bad. Three oranges are drawn at random without replacement. The box is approved only if all three are good.
To find: The probability that the box is approved.
Plan: (1) Name an event for each draw being good. (2) Find the probability of each draw, given the earlier ones were good. (3) Multiply them. (4) Check by counting with combinations.

Basics you need

1Classical probability

$P(E) = \frac{n(E)}{n(S)}$ when the outcomes are equally likely. "Randomly selected" tells us every orange in the box is equally likely to be drawn. Basics: P(E) = n(E)/n(S)

2Conditional probability for a draw

$P(F|E)$ is found by looking at what is left in the box once $E$ has happened. Lesson: conditional probability

3Multiplication rule for three events

$P(E\cap F\cap G) = P(E)\,P(F|E)\,P(G|EF)$ ⁠, where $EF$ means $E\cap F$ (Section 13.3). Lesson: the multiplication rule

Solution

Name the events

Let $E$ be the event "the first orange drawn is good", $F$ the event "the second orange drawn is good" and $G$ the event "the third orange drawn is good". The box is approved exactly when $E\cap F\cap G$ happens.

"Otherwise, it is rejected" means a single bad orange among the three is enough to reject the box. So "approved" is the one event "all three good", and that is the intersection of the three single-draw events.

An intersection of events means they all happen. Basics: events as sets
The first draw: $P(E)$

The box has 15 oranges, 12 of them good:
$P(E) = \frac{12}{15} = \frac{4}{5}.$
Each of the 15 oranges is equally likely to be the first one picked, and 12 of them are favourable.

Classical definition of probability. Basics: classical probability
The second draw, given the first was good: $P(F|E)$

One good orange has been taken out and not replaced. Left: 14 oranges, of which $12 - 1 = 11$ are good:
$P(F|E) = \frac{11}{14}.$
Because we condition on $E$ ⁠, we count in the box as it is after a good orange has gone. The bad oranges are all still there (3 of them), so $11 + 3 = 14$ is the new total.

Conditional probability as the probability in the reduced sample space (Section 13.2). Lesson: conditional probability
The third draw, given the first two were good: $P(G|EF)$

Two good oranges have now gone. Left: 13 oranges, of which $12 - 2 = 10$ are good:
$P(G|EF) = \frac{10}{13}.$
The condition is "the first two were both good", so we use the box after two good oranges have been removed: 10 good and 3 bad.

$P(G|EF)$ is the probability of $G$ given that $E$ and $F$ have both happened. Lesson: the multiplication rule
Multiply the three probabilities

By the multiplication rule of probability for three events,
$\begin{aligned} &P(E\cap F\cap G) \\ &= P(E)\,P(F|E)\,P(G|EF) \\ &= \frac{12}{15}\times\frac{11}{14}\times\frac{10}{13} \\ &= \frac{1320}{2730} \\ &= \frac{44}{91}. \end{aligned}$
To simplify: $1320 = 30\times 44$ and $2730 = 30\times 91$ ⁠. Since $44 = 2^2\times 11$ and $91 = 7\times 13$ share no factor, $\frac{44}{91}$ is in lowest terms.

The only path to "approved" is good (G), good, good. Each green probability is read from the box as the earlier good draws left it: $\frac{12}{15}$ ⁠, then $\frac{11}{14}$ ⁠, then $\frac{10}{13}$ ⁠. A bad orange (B) at any draw ends in rejection.

The rule chains the conditional probabilities: first good, then good given the first was good, then good given the first two were good. On the diagram that is multiplying along the single green path. The book uses the same three-event rule for the kings and the ace in Example 9.

Multiplication rule for more than two events, Section 13.3: $P(E\cap F\cap G) = P(E)\,P(F|E)\,P(G|EF)$ ⁠. Lesson: Examples 8 and 9
Check by counting with combinations

The three oranges form a random set of 3 from 15. All $^{15}C_3$ sets are equally likely, and $^{12}C_3$ of them contain only good oranges:
$\begin{aligned} P &= \frac{^{12}C_3}{^{15}C_3} = \frac{220}{455} \\ &= \frac{44\times 5}{91\times 5} = \frac{44}{91}. \end{aligned}$
Here $^{12}C_3 = \frac{12\times 11\times 10}{6} = 220$ and $^{15}C_3 = \frac{15\times 14\times 13}{6} = 455$ ⁠. The same answer.

Counting sets ignores the order of the draws, while the multiplication rule follows the order. Both describe the same random choice, so they must agree, and they do.

$^{n}C_3 = \frac{n(n-1)(n-2)}{6}$ (Class XI). Basics: counting

Result

P(box approved) $= \frac{12}{15}\times\frac{11}{14}\times\frac{10}{13} = \frac{44}{91}$ ⁠.

(Textbook answer: $\frac{44}{91}$ ⁠.)

Check it with numbers

Count ordered draws: $15\times 14\times 13 = 2730$ ways to draw three oranges in order, and $12\times 11\times 10 = 1320$ of them are all good. So the probability is $\frac{1320}{2730} \approx 0.4835$ ⁠, and $\frac{44}{91} \approx 0.4835$ ⁠. A box with 3 bad oranges in 15 is approved a little less than half of the time.

Common mistakes

Using $\left(\frac{12}{15}\right)^3$ ⁠. That is $\frac{64}{125}$ ⁠, the answer with replacement. Without replacement both the good count and the total drop by one after each good draw.
Lowering only the total. $\frac{12}{15}\times\frac{12}{14}\times\frac{12}{13}$ forgets that each orange removed was a good one.
Answering with the probability of rejection. Rejection is the complement: $1 - \frac{44}{91} = \frac{47}{91}$ ⁠. The question asks for approval.

What to take away

For draws without replacement, multiply the conditional probabilities: $P(E)\,P(F|E)\,P(G|EF)$ ⁠.
After each draw, update both the number of favourable items and the total.
$\frac{^{12}C_3}{^{15}C_3}$ gives the same answer and is a quick independent check.

Every fact used, and where it comes from

Fact	Used in	Source
$P(E) = \frac{n(E)}{n(S)}$ for equally likely outcomes	Step 2, Check	Class XI, Probability (recalled in Section 13.1)
Conditional probability in the reduced sample space	Steps 3, 4	Section 13.2, Definition 1 (Lesson: conditional probability)
$P(E\cap F\cap G) = P(E)\,P(F\|E)\,P(G\|EF)$	Step 5	Multiplication rule for more than two events, Section 13.3
Three draws without replacement, same method	Step 5	Example 9, Section 13.3
$^{n}C_3 = \frac{n(n-1)(n-2)}{6}$	Step 6	Class XI, Permutations and Combinations

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Exercise 13.2 · Q3: Box of oranges approved: three good ones drawn

Basics you need

Solution

Name the events

The first draw: P(E)P(E)P(E)

The second draw, given the first was good: P(F∣E)P(F|E)P(F∣E)

The third draw, given the first two were good: P(G∣EF)P(G|EF)P(G∣EF)

Multiply the three probabilities

Check by counting with combinations

Every fact used, and where it comes from

The first draw: $P(E)$

The second draw, given the first was good: $P(F|E)$

The third draw, given the first two were good: $P(G|EF)$