Bernoulli,Binomial,Geometric,Poisson,Uniform,Exponential examples

1. SEREN NALKIRAN MİYASE KÜBRA ERKAN
120309044 120309024
BURCU ÖZTÜRK ESRA ŞAVKI
120309048 120309066
Topic : QUESTİONS

2. QUESTİON OF BERNOULLİ
 A fair coin is tossed. Let the variable x take values 1 and 0 according to as the toss
results in ‘Head‘ or ‘Tail’. Then, X is a Bernoulli variable with parameter p=1/2. Here, X
denotes the number of heads obtained in the toss. Probability of success = 1/2 and the
probability of failure = 1/2.
 In a single throw of a dice, the outcome "5" is called a success and any other outcome
is called a failure, then the successive throws of a dice will contain Bernoulli trials.The
probability of success = 1/6 and the probability of failure = 5/6.
 Solved Examples
 Question 1: Find the probability of getting a head in a single toss of a coin.
Solution:
A fair coin is tossed. Let the variable x take values 1 and 0 according to as the toss
results in ‘Head ‘ or ‘Tail’. Then X is a Bernoulli variable with parameter p=1/2.
Here, X denotes the number of heads obtained in the toss.
=> Probability of success = 1/2 and the probability of failure = 1/2.
 Question 2: Find the probability of getting 5 in a single throw of a dice.
Solution:
In a single throw of a dice, the outcome "5" is called a success and any other outcome
is called a failure, then the successive throws of a dice will contain Bernoulli trials.
The probability of success = 1/6 and the probability of failure = 5/6.

3.  Question 3: On a multiple choice examination with three possible
answers for each of the five questions, what is the probability that a
candidate get 4 or more correct answers just by guessing ?
Solution:
Let x is the number of correct answers, then x has binmonial
distribution with
n = 5, p = 13 and q = 1 - p = 1 - 1/3 = 2/3
Required probability = P(x ≥ 4) = P(x = 4) + P(x = 5)
= 5C4 (2/3)5−4 (1/3)4 + 5C5 (2/3)5−5(1/3)5
= 5 * 2/3 * 1/81 + 1 * 1 * 1/243
= 10+1/243
= 11/243
QUESTİON OF BİNOMİAL

4. QUESTİON OF GEOMETRİC
 Question 4 : Four roads start from a junction. Only one of them leads to a mall.
The remaining roads ultimately lead back to the starting point. A person not
familiar with these roads wants to try the different roads one by one to reach the
mall. What is the probability that his second attempt will be correct?
Solution:
This is an experiment of probability. Here the person has a random probability
of 14, for choosing the correct road. The problem statement suggests the
probability distribution to be geometric as the requirement is a success in the
second trial.
The probability of success is given by the geometric distribution formula,
P(X = k) = p * (1 – p)k – 1, where, p = 1/4 = 0.25 and k = 2
Therefore, the required probability P(X = 2) = 0.25*(1 – 0.25)2 – 1
= 0.25 * (0.75)
= 0.1875
= 18.75%

5. QUESTIONS OF POISSON
Question 5 :Telephone calls arrive at an exchange according to Poisson process with rate λ
= 2/min. Calculate the probability that exactly 2 calls will be received during each of the first
5 min of the hour.
Solution:
Step 1:
Let N be the number of calls received during a 1 min period.
=> P(N = 2) = e−2*4/2!
= 2e−2
Step 2:
Let M be the number of minute, among the 5 min considered, during which exactly 2 calls
will be received. M follows a binomial distribution with parameters n = 5 and p = 2e−2.
P(M = 5) = C(5, 5)(2e−2)5(1−2e−2)5−5
= 32 * e−10
≈ 0.00145.

6. QUESTİON OF UNİFORM
Question 6: Suppose you are conducting a quiz and post a question to the
audience of 20 competitors. The time allowed to answer the question is 30
seconds. How many persons are likely to respond within 5 seconds?
(Normally, the competitors are required to click a button of the correct choice
and the winner is chosen on the basis of first click).
Solution:
Step 1:
The interval of the probability distribution in seconds is [0, 30].
=> The probability density is = 1/30–0=1/30.
Step 2:
The requirement is how many will respond in 5 seconds. That is, the sub
interval of the successful event is [0, 5].
Now the probability P ( x < 5) is the ratio of the widths of these two intervals.
=> 5/30=1/6.
Since there are 20 competitors, the number of competitors likely to respond
in 5 seconds is (1/6)(20) ≈ 3.

7. QUESTİON OF EXPONENTİAL
Question: Let take a customer who goes for shopping and lets suppose the time he spends in the shop is
exponentially distributed with a mean value 15 minutes. Then what will be the probability that the customer will
spend more than 20 minutes in shopping? What is the probability that the customer will spend more than 20
minutes in the bank given that he is still in the shop after 15 minutes?
Solution:
Here mean is given as 15 minutes
Now the mean of an exponential distribution is 1λ
So 1/λ = 10
So λ = 1/10
So P (X > 20) = e−20λ
= e−20/15
= e−4/3
= 0.26
For the next part we need to find the probability that the customer will spend more than 20 minutes in the bank
given that he is still in the shop after 15 minutes.
So P(X > 20) at X > 15 = P(X > 5)
= e−1/2
= 0.604