Chapter 13 · Probability

Miscellaneous · Q8: A $2 \times 2$ determinant of 0s and 1s is positive, $P = \frac{3}{16}$

Miscellaneous ExerciseIndependent events · Section 13.4

The question

If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability $\frac12$ ⁠).

Exam answer

Let the determinant be $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$ ⁠, where each of $a, b, c, d$ is 0 or 1.

The entries are independent, each 0 or 1 with probability $\frac12$ ⁠. So there are $2^4 = 16$ determinants, each with probability $\left(\frac12\right)^4 = \frac{1}{16}$ ⁠.

Let $E$ ⁠: the determinant is positive. Since $ad$ and $bc$ are each 0 or 1, $ad - bc > 0$ only when $ad = 1$ and $bc = 0$ ⁠.

$ad = 1$ only when $a = d = 1$ ⁠: 1 way. $bc = 0$ when $(b, c) = (0, 0), (0, 1)$ or $(1, 0)$ ⁠: 3 ways.

So $n(E) = 1 \times 3 = 3$ ⁠: $(a, b, c, d) = (1, 0, 0, 1)$ ⁠, $(1, 0, 1, 1)$ and $(1, 1, 0, 1)$ ⁠.

$\therefore\ P(E) = \dfrac{n(E)}{n(S)} = \dfrac{3}{16}$ ⁠.

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

The one idea this problem is built on

A second order determinant is $ad - bc$ ⁠. With entries 0 or 1, both products $ad$ and $bc$ are 0 or 1, so the determinant is positive only in the one way " $ad = 1$ and $bc = 0$ ". Counting the 16 equally likely determinants (or multiplying independent probabilities) gives $\frac{3}{16}$ ⁠.

Given, to find, and the plan

Given: A determinant $\begin{vmatrix} a & b \\ c & d \end{vmatrix}$ whose entries are chosen independently, each equal to 0 or 1 with probability $\frac12$ ⁠.
To find: P(the value of the determinant is positive).
Plan: (1) Write the value $ad - bc$ and its possible values. (2) Set up the 16 equally likely cases. (3) Decide when the value is positive. (4) Count. (5) Confirm by independence.

Basics you need

1A determinant of order 2

$\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$ (Chapter 4, Determinants).

2Multiply the choices

4 entries, 2 choices each: $2 \times 2 \times 2 \times 2 = 16$ determinants. Basics: counting

3Independent events

For independent entries the probability of a combination is the product, so each of the 16 cases has probability $\left(\frac12\right)^4$ (Section 13.4). Lesson: independent events

4Classical probability

$P(E) = \frac{n(E)}{n(S)}$ for equally likely outcomes. Basics: P(E) = n(E)/n(S)

Solution

The value of the determinant

Write the determinant as
$\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc,$
where each of $a, b, c, d$ is 0 or 1. A product of two such numbers is 1 only when both are 1, and 0 otherwise. So $ad$ and $bc$ are each 0 or 1, and the value $ad - bc$ is $-1$ ⁠, $0$ or $1$ ⁠.

Turning the determinant into a number tells us exactly what "positive" means here: the value must be 1.

Determinant of order 2 (Chapter 4).
The 16 equally likely determinants

Each entry has 2 possible values, so there are $2^4 = 16$ possible determinants. The entries are chosen independently, each value with probability $\frac12$ ⁠, so each determinant has probability
$\frac12 \times \frac12 \times \frac12 \times \frac12 = \frac{1}{16}.$
Independence lets us multiply the four probabilities. All 16 cases come out equal, so we may count: $P(E) = \frac{n(E)}{16}$ ⁠.

Mutually independent events multiply (Section 13.4, Remark (iv), extended to four events). Lesson: independent events
When is the value positive?

Let $E$ ⁠: the determinant is positive. As $ad$ and $bc$ are each 0 or 1,
$ad - bc > 0 \iff ad = 1 \text{ and } bc = 0.$
$ad = 1$ only when $a = 1$ and $d = 1$ ⁠. $bc = 0$ when at least one of $b, c$ is 0: $(b, c) = (0, 0), (0, 1)$ or $(1, 0)$ ⁠.

The other combinations give $1 - 1 = 0$ ⁠, $0 - 0 = 0$ or $0 - 1 = -1$ ⁠, none of them positive.

An event described in words becomes a set of outcomes. Basics: events as sets
Count the favourable cases

There is 1 choice for $(a, d)$ and 3 choices for $(b, c)$ ⁠, so $n(E) = 1 \times 3 = 3$ ⁠. The three positive determinants have $(a, b, c, d) = (1, 0, 0, 1)$ ⁠, $(1, 0, 1, 1)$ and $(1, 1, 0, 1)$ ⁠. So
$P(E) = \frac{n(E)}{n(S)} = \frac{3}{16}.$

All 16 determinants. The rows fix $a$ and $d$ ⁠, the columns fix $b$ and $c$ ⁠, and each cell shows $ad - bc$ ⁠. Only the 3 shaded cells in the bottom row have the value 1.

The choice of $(a, d)$ and the choice of $(b, c)$ are separate, so the counts multiply.

Multiply the choices. Basics: counting
The same answer by independence

$ad = 1$ needs $a = 1$ and $d = 1$ ⁠, so $P(ad = 1) = \frac12 \times \frac12 = \frac14$ ⁠. $bc = 0$ is the complement of " $b = 1$ and $c = 1$ ", so $P(bc = 0) = 1 - \frac14 = \frac34$ ⁠. The two events use different entries, so they are independent, and
$P(E) = \frac14 \times \frac34 = \frac{3}{16}.$
" $ad = 1$ " depends only on $a$ and $d$ ⁠, and " $bc = 0$ " only on $b$ and $c$ ⁠; since all four entries are independent, the two events are independent too.

$P(E \cap F) = P(E)\,P(F)$ for independent events (Section 13.4, Definition 3). Lesson: independent events

Result

The probability that the determinant is positive is $\frac{3}{16}$ ⁠. (Textbook answer: $\frac{3}{16}$ ⁠.)

Check it with numbers

Of the 16 determinants, 3 have value 1, 3 have value $-1$ (the mirror cases with $ad = 0$ and $b = c = 1$ ⁠), and the other 10 have value 0: $3 + 3 + 10 = 16$ ⁠. So $P(E) = \frac{3}{16} = 0.1875$ ⁠. The verify script lists all 16 and computes each determinant.

Common mistakes

Counting only $(1, 0, 0, 1)$ ⁠. $bc$ is 0 whenever at least one of $b, c$ is 0; there are 3 such pairs, not 1.
Including $a = b = c = d = 1$ ⁠. Its value is $1 - 1 = 0$ ⁠, which is not positive.
Answering $\frac13$ ⁠. The three values $-1, 0, 1$ are not equally likely; count the 16 cases.

What to take away

Turn the condition into numbers first: here "positive" means $ad = 1$ and $bc = 0$ ⁠.
Independent entries with probability $\frac12$ each make all $2^n$ cases equally likely, so counting works.
Events that depend on separate independent entries are independent: $\frac14 \times \frac34$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
$\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$	Step 1	Chapter 4, Determinants
Multiply the choices: $2^4 = 16$	Steps 2, 4	Counting, Class XI (Basics: counting)
Mutually independent events multiply	Steps 2, 5	Section 13.4, Definition 3 and Remark (iv) (Lesson: independent events)
$P(E) = \frac{n(E)}{n(S)}$ for equally likely outcomes	Step 4	Class XI, Probability (Basics: classical probability)
$P(E') = 1 - P(E)$	Step 5	Class XI axioms (Basics: rules of probability)

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Miscellaneous · Q8: A 2×22 \times 22×2 determinant of 0s and 1s is positive, P=316P = \frac{3}{16}P=163​

Basics you need

Solution

The value of the determinant

The 16 equally likely determinants

When is the value positive?

Count the favourable cases

The same answer by independence

Every fact used, and where it comes from

Miscellaneous · Q8: A $2 \times 2$ determinant of 0s and 1s is positive, $P = \frac{3}{16}$