Banyak sekali macam distribusi peluang diskrit, berikut ini adalah dua di antaranya.

1. DISTRIBUSI
BINOMIAL

2. BERNOULLI PROCESS
 The experiment consists of n repeated
trials.
 Each trial results in an outcome that
may be classified as a success or a
failure.
 The probability of success, denoted by
p, remains constant from trial to trial.
 The repeated trials are independent.

3. BINOMIAL
DISTRIBUTION
A Bernoulli trial can result in a success with
probability p and a failure with probability q
= 1-p. Then the probability distribution of the
binomial random variable X, the number of
successes in n independent trials, is
  xnx
xn qpCpnxb 
,;

4. CONTOH SOAL
DISTRIBUSI BINOMIAL (1)
 The probability that a certain kind of
component will survive a given shock test is
¾. Find the probability that exactly 2 of the
next 4 components tested survive!
128
27
4
3
!2!2
!4
4
1
4
3
4
3
,4;2 4
222
24 

















Cb

5. CONTOH SOAL
DISTRIBUSI BINOMIAL (2)
 The probability that a patient recovers from
a rare blood disease is 0.4. If 15 people are
known to have contracted this disease, what
is the probability that exactly 5 survive?
  1859.06.04.04.0,15;5 105
515  Cb

6. LATIHAN (1)
 Sebuah produsen obat batuk memberikan
pernyataan bahwa obat batuknya 80%
efektif dalam menyembuhkan penyakit
batuk. Apabila 7 orang dengan batuk serupa
diberikan obat dari produsen itu, tentukan
peluangnya (a) tepat 3 orang di antaranya
sembuh (b) tepat 5 orang di antaranya
sembuh (c) semuanya sembuh

7. LATIHAN (2)
 Seorang pemain basket memiliki peluang
0,7 untuk memasukkan bola ke dalam ring
basket. Apabila ia melakukan 10 lemparan
berturutan, tentukan peluang (a) tepat 6
bola berhasil masuk ring (b) tepat 3 bola
berhasil masuk ring (c) tak kurang dari 8
bola masuk ring (d) tak ada bola yang
masuk

8. LATIHAN (3)
 In a certain city district the need for money
to buy drugs is given as the reason for 75%
of all thefts. Find the probability that among
the next 5 theft cases reported in this
district, (a) exactly 2 resulted from the need
for money to buy drugs; (b) at most 3
resulted from the need for money to buy
drugs.

9. LATIHAN (4)
 A traffic control engineer reports that
75% of the vehicles passing through a
checkpoint are from within the state.
What is the probability that more than
2 of the next 9 vehicles are from out of
the state?

10. DISTRIBUSI
POISSON

11. POISSON PROBABILITY
EXPERIMENT
 The random variable is the number of
times some event occurs during a
defined interval.
 The probability of the event is
proportional to the size of the interval.
 The intervals do not overlap and are
independent.

12. POISSON
DISTRIBUTION
 The probability distribution of the Poisson
random variable X, representing the number
of outcomes occuring in a given time interval
or specified region denoted by t, is given by:
    ,...3,2,1,0;
!
; 

x
x
te
txp
xt



e  2,718281828

13. CONTOH SOAL
DISTRIBUSI POISSON (1)
 Rata-rata banyaknya nasabah yang masuk ke
dalam antrian bagian teller suatu bank setiap
menitnya adalah 2. Tentukan peluang dalam 1
menit datang 3 nasabah ke dalam antrian bagian
teller tersebut!
  .1804.0
!3
2
2;3
32




e
p

14. CONTOH SOAL
DISTRIBUSI POISSON (2)
Seorang sekretaris rata-rata menerima panggilan telepon
sebanyak 3 buah dalam setiap 20 menit. Tentukan peluang
dalam 1 jam berikutnya ia menerima 7 buah panggilan
telepon.
Jawab:
λ = 0,15/menit, t = 60 menit
λt = 0,15 . 60 = 9

15. CONTOH SOAL
DISTRIBUSI POISSON (3)
 Banyaknya kata yang salah ejaan dalam suatu surat kabar
adalah 3 dalam tiap 4 halaman. Tentukan peluang dalam 10
halaman surat kabar tersebut terdapat kurang dari 4 kata
salah ejaan.
 λ = 0,75/halaman t = 10 halaman λt = 0,75 . 10 = 7,5
 P[X<4] = p(0;7,5) + p(1;7,5) + p(2;7,5) + p(3;7,5)
= 0,0006 + 0,0041 + 0,0156 + 0,0389 = 0,0592.

16. CONTOH SOAL
DISTRIBUSI POISSON (4)
 Pada contoh soal sebelumnya, tentukan peluang dalam 10
halaman surat kabat tersebut terdapat lebih dari 3 kata salah
ejaan.
 P[X>3] = p(4;7,5) + p(5;7,5) + p(6;7,5) + p(7;7,5) + ...
= 1 – [p(0;7,5) + p(1;7,5) + p(2;7,5) + p(3;7,5)]
= 1 – 0,0592 = 0,9408.