Practice: conditional probability and independence

Probability, grade 11 (5 units) Lesson 8 of 16Practice

Practice: conditional probability and independence

1

In a survey 600 adults were asked whether they use a smartwatch:

	Use	Do not use	Total
Aged 18–30	90	110	200
Aged 31–50	75	175	250
Aged 51 and over	30	120	150
Total	195	405	600

An adult is chosen at random.

a

What is the probability that the adult uses a smartwatch?

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

b

The adult is known to be aged 18–30. What is the probability that the adult uses a smartwatch?

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

c

The adult is known to use a smartwatch. What is the probability that the adult is aged 51 or over?

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

d

Are “the adult is aged 51 or over” and “the adult uses a smartwatch” independent events? Explain.

2

35% of a library’s readers are students.
60% of the students borrowed an e-book last month.
20% of the readers who are not students borrowed an e-book last month.
Write the area of each rectangle in the area model, and mark the rectangles of the readers who borrowed an e-book.

a

What is the probability that a reader chosen at random borrowed an e-book?

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

b

A reader is known to have borrowed an e-book. What is the probability that the reader is a student?

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

c

A reader is known not to have borrowed an e-book. What is the probability that the reader is not a student?

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

3

$\displaystyle {P(A)}={0.6}$ , $\displaystyle {P(B/A)}={0.5}$ , $\displaystyle {P(B)}={0.42}$ .

a

Write $\displaystyle {P(B/\overline{A})}={x}$ , and find $\displaystyle x$ . A two-way table of probabilities helps.

b

Work out $\displaystyle P(A/B)$ .

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

4

$\displaystyle A$ and $\displaystyle B$ are independent events.
$\displaystyle {P(A)}={0.3}$ , $\displaystyle {P(\overline{B})}={0.2}$ .

a

Work out $\displaystyle P(A\cap B)$ .

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

b

Work out $\displaystyle P(A\cup B)$ .

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

c

What is the probability that exactly one of the two events happens?

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

5

$\displaystyle A$ and $\displaystyle B$ are independent events.
The probability of $\displaystyle A$ is twice the probability of $\displaystyle B$ .
$\displaystyle {P(A\cap B)}={0.18}$ .

a

Write $\displaystyle {P(B)}={x}$ , and find $\displaystyle x$ .

b

Work out $\displaystyle P(A\cup B)$ .

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

6

$\displaystyle P(A)>0$ and $\displaystyle P(B)>0$ .
It is known that $\displaystyle P(A/B)>P(A)$ .
Prove that $\displaystyle P(B/A)>P(B)$ . Use the definition of conditional probability.

Formulas: probability, grade 11 (5 units)

The rules of probability

Equally likely outcomes – $\displaystyle k$ the outcomes in the event $\displaystyle A$ , $\displaystyle n$ the outcomes in the sample space
$\displaystyle P(A)=\frac{k}{n}$
Every event; $\displaystyle {P(\emptyset)}={0}$ , and $\displaystyle {P(U)}={1}$ for the sample space $\displaystyle U$
$\displaystyle 0\le P(A)\le1$
An event contained in another event
$\displaystyle P(A)\le P(B)$ $\displaystyle A\subseteq B$
The complement $\displaystyle \overline{A}$ – $\displaystyle A$ does not happen
$\displaystyle P(\overline{A})$ $\displaystyle =$ $\displaystyle 1-P(A)$
The union – $\displaystyle A$ or $\displaystyle B$ (or both); $\displaystyle A\cap B$ – the intersection, both $\displaystyle A$ and $\displaystyle B$
$\displaystyle P(A\cup B)$ $\displaystyle =$ $\displaystyle P(A)+P(B)-P(A\cap B)$
Disjoint events – no two of them have an outcome in common
$\displaystyle P(A_1\cup A_2\cup\dots\cup A_n)=P(A_1)+P(A_2)+\dots+P(A_n)$

Conditional probability and independence

The probability of $\displaystyle A$ given $\displaystyle B$
$\displaystyle P(A/B)=\frac{P(A\cap B)}{P(B)}$
A path in a tree – the product of the probabilities on its branches
$\displaystyle P(A\cap B)$ $\displaystyle =$ $\displaystyle P(B)\cdot P(A/B)$
Independent events
$\displaystyle P(A\cap B)$ $\displaystyle =$ $\displaystyle P(A)\cdot P(B)$
Independent events – knowing that $\displaystyle B$ happened does not change the probability of $\displaystyle A$
$\displaystyle P(A/B)$ $\displaystyle =$ $\displaystyle P(A)$
The law of total probability – the paths that end in $\displaystyle A$ , added up
$\displaystyle P(A)=P(B)\cdot P(A/B)+P(\overline{B})\cdot P(A/\overline{B})$

Counting and Bernoulli's formula

$\displaystyle n$ factorial; $\displaystyle {0!}={1}$
$\displaystyle n!=1\cdot2\cdot3\cdot\dots\cdot n$
Ordered selections – $\displaystyle k$ different objects out of $\displaystyle n$ , one after another
$\displaystyle n\cdot(n-1)\cdot\dots\cdot(n-k+1)=\frac{n!}{(n-k)!}$
The binomial coefficient – the ways to choose $\displaystyle k$ objects out of $\displaystyle n$ , order not counted
$\displaystyle \binom{n}{k}=\frac{n!}{k!(n-k)!}$
Bernoulli’s formula – the probability of $\displaystyle k$ successes in $\displaystyle n$ independent trials, with the probability of success $\displaystyle p$ in each
$\displaystyle P_n(k)=\binom{n}{k}p^k(1-p)^{n-k}$
At least one success in $\displaystyle n$ trials – by the complement
$\displaystyle 1-P_n(0)=1-(1-p)^n$