Probability Theory

Assigning Probability

Let

S

be the sample space of an experiment with a finite or countable number of outcomes. We assign a probability

p(s)

to each outcome

s

. We require that two conditions be met:

Misalkan $S$ menjadi ruang sampel percobaan dengan jumlah hasil yang terbatas atau dapat dihitung. Kami menetapkan $p$ probabilitas untuk setiap hasil. Kami mengharuskan dua kondisi terpenuhi:

~~~ (i) ~~~ 0 \leq p(s) \leq 1

for each

s \in S

and

~~~ (ii) ~~~ \sum_{s \in S} p(s) = 1

Example

What probabilities should we assign to the outcomes H (heads) and T (tails) when a fair coin is flipped? What probabilities should be assigned to these outcomes when the coin is biased so that heads comes up twice as often as tails?

Berapa Probabilitas yang harus kita berikan pada hasil H (kepala) dan T (ekor) ketika koin yang adil dibalik? Peluang apa yang harus diberikan untuk hasil ini ketika koin bias sehingga kepala muncul dua kali lebih sering dari pada ekor?

Solution:

For a fair coin, the probability that heads comes up when the coin is flipped equals the probability that tails comes up, so the outcomes are equally likely.
Consequently, we assign the probability

\frac{1}{2}

to each of the two possible outcomes, that is,

p(H) = p(T) = \frac{1}{2}

.
For the biased coin we have

~~~~~~~~~~ p(H) = 2p(T)

.
Because

~~~~~~~~~~ p(H) + p(T) = 1

,
it follows that

~~~~~~~~~~ 2p(T) + p(T) = 3p(T) = 1

.
We conclude that\

p(T) = \frac{1}{3}

and

p(H) = \frac{2}{3}

Uniform Distribution

Suppose that

S

is a set with

n

elements. The uniform distribution assigns the probability

\frac{1}{n}

to each element of

S

Misalkan $S$ adalah himpunan dengan $n$ elemen. Distribusi seragam memberikan probabilitas $\frac{1}{n}$ untuk setiap elemen $S$ .

Conditional Probability

Let

E

and

F

be events with

p(F) > 0

. The conditional probability of

E

given

F

, denoted by

p(E | F)

, is defined as

p(E | F) = \dfrac{p(E \cap F)}{p(F)}

Example

A bit string of length four is generated at random so that each of the 16 bit strings of length four is equally likely. What is the probability that it contains at least two consecutive 0s, given that its first bit is a 0? (We assume that 0 bits and 1 bits are equally likely.)

Sebuah string bit dengan panjang empat dihasilkan secara acak sehingga masing-masing dari 16 bit string dengan panjang empat kemungkinannya sama. Berapa probabilitasnya mengandung setidaknya dua 0 berturut-turut, mengingat bit pertamanya adalah 0? (Kami berasumsi bahwa 0 bit dan 1 bit kemungkinannya sama.)

Solution:

Let

E

be the event that a bit string of length four contains at least two consecutive 0s, and let

F

be the event that the first bit of a bit string of length four is a 0.

Misalkan E adalah kejadian di mana sebuah string bit dengan panjang empat berisi setidaknya dua 0 berturut-turut, dan misalkan F adalah kejadian bahwa bit pertama dari string bit dengan panjang empat adalah 0.

The probability that a bit string of length four has at least two consecutive 0s, given that its first bit is a 0, equals

~~~~~~~~~~ p(E | F) = \dfrac{p(E \cap F)}{p(F)}

.
Because

E \cap F = {0000, 0001, 0010, 0111, 0100}

,
we see that

p(E \cap F) = \frac{5}{16}

.
Because there are eight bit strings of length four that start with a 0, we have

p(F) = \frac{8}{16} = \frac{1}{2}

. Consequently,

~~~~~~~~~~ p(E | F) = \dfrac{5/16}{1/2} = \dfrac{5}{8}

What is the conditional probability that a family with two children has two boys, given they have at least one boy? Assume that each of the possibilities BB, BG, GB, and GG is equally likely, where B represents a boy and G represents a girl. (Note that BG represents a family with an older boy and a younger girl while GB represents a family with an older girl and a younger boy.)

Berapa probabilitas bersyarat bahwa sebuah keluarga dengan dua anak memiliki dua anak laki-laki, mengingat mereka memiliki setidaknya satu anak laki-laki? Asumsikan bahwa masing-masing kemungkinan BB, BG, GB, dan GG memiliki kemungkinan yang sama, di mana B mewakili laki-laki dan G mewakili perempuan. (Perhatikan bahwa BG mewakili sebuah keluarga dengan seorang anak laki-laki yang lebih tua dan seorang anak perempuan yang lebih muda sementara GB mewakili sebuah keluarga dengan seorang anak perempuan yang lebih tua dan seorang anak laki-laki yang lebih muda.)

Solution:

Let

E

be the event that a family with two children has two boys, and let

F

be the event that a family with two children has at least one boy.
It follows that

E = {BB}, F = {BB, BG, GB}

, and

E ∩ F = {BB}

.
Because the four possibilities are equally likely, it follows that

p(F) = \frac{3}{4}

and

p(E \cup F) = \frac{1}{4}

. We conclude that

~~~~~~~~~~ p(E | F) = \dfrac{p(E \cap F)}{p(F)} = \dfrac{1/4}{3/4} = \dfrac{1}{3}

Independence

Suppose a coin is flipped three times, does knowing that the first flip comes up tails (event

F

) alter the probability that tails comes up an odd number of times (event

E

Misalkan sebuah koin dibalik tiga kali, bagaimana mengetahui bahwa lemparan pertama muncul ekor (peristiwa $F$ ) mengubah probabilitas bahwa ekor muncul dalam jumlah ganjil (peristiwa $E$ )?

In other words, is it the case that

p(E | F) = p(E)

? This equality is valid for the events

E

and

F

, because

p(E | F) = \frac{1}{2}

and

p(E) = \frac{1}{2}

Because

p(E | F) = \dfrac{p(E \cap F)}{p(F)}

, asking whether

p(E | F) = p(E)

is the same as asking whether

p(E \cap F) = p(E)p(F)

The events

E

and

F

are independent if and only if

p(E \cap F) = p(E)p(F)

Example

Suppose

E

is the event that a randomly generated bit string of length four begins with a 1 and

F

is the event that this bit string contains an even number of 1s. Are

E

and

F

independent, if the 16 bit strings of length four are equally likely?Solution:

There are eight bit strings of length four that begin with a one: 1000, 1001, 1010, 1011, 1100, 1101, 1110, and 1111.
There are also eight bit strings of length four that contain an even number of ones: 0000, 0011, 0101, 0110, 1001, 1010, 1100, 1111.
Because there are 16 bit strings of length four, it follows that

~~~~~~~~~~ p(E) = p(F) = \dfrac{8}{16} = \dfrac{1}{2}

.
Because

E \cap F = {1111, 1100, 1010, 1001}

, we see that

~~~~~~~~~~ p(E \cap F) = \dfrac{4}{16} = \dfrac{1}{4}

.
Because

~~~~~~~~~~ p(E \cap F) = \dfrac{1}{4} = \dfrac{1}{2} \times \dfrac{1}{2} = p(E)p(F)

,
we conclude that

E

and

F

are independent.

Persiapan Belajar

Pertemuan 1

Pertemuan 2

Pertemuan 3

Pertemuan 4

Pertemuan 6

Pertemuan 7

UTS

Pertemuan 9

Probability Theory

Assigning Probability

Example

Uniform Distribution

Conditional Probability

Example

Independence

Example

Persiapan Belajar

Pertemuan 1

Pertemuan 2

Pertemuan 3

Pertemuan 4

Pertemuan 6

Pertemuan 7

UTS

Pertemuan 9

​Assigning Probability

​Example

​Uniform Distribution

​Conditional Probability

​Example

​Independence

​Example

Assigning Probability

Example

Uniform Distribution

Conditional Probability

Example

Independence

Example