Transcript Lesson One Bernoulli and Binomial Distributions File

Binomial
Distribution
Bernoulli Trials
Repeated identical trials are called Bernoulli trials if:
1. There are two possible outcomes for each trial, denoted
by s (for success) and f (for failure).
2. The trials are independent.
3. The probability of success remains the same from trial to
trial. We denote the probability of success by the letter p.
4. The random variable can only take the values of 0 or 1.
X
B(1, p)
or
1 x
P( X  x)  p (1  p)
for x  0, 1
x
A game is played by rolling a die. If the die comes up with a 1 or
2, you win the game. Otherwise you lose the game. What is the
probability that you win the game?
Answer: 0.333
To find the mean, or expectation, of a Bernoulli Distribution we use:
E( X )     x  P( X  x)
Therefore:
E ( X )  0   p 0 (1  p)1   1 ( p1 (1  p) 0 )  p
To find the variance of a Bernoulli Distribution we use:
Var ( X )  E ( X 2 )  ( E ( X ))2
Therefore:
Var ( x)  [0   p 0 (1  p)1   1 ( p1 (1  p)0 )]  p 2
Var ( X )  p  p  p(1  p)
2
Mean of a Bernoulli Distribution:
A Bernoulli Distribution with parameter p has a mean of p.
Variance of a Bernoulli Distribution:
A Bernoulli Distribution with parameter p has a variance of p(1 – p).
A coin is flipped three times. What are the possible combinations of
getting 0 heads? 1 head? 2 heads? 3 heads?
Our possibilities are:
0 heads:
TTT
1 head:
TTH
HTT
THT
2 heads:
HHT
HTH
THH
3 heads:
HHH
A coin is flipped four times. What are the possible combinations of
getting 0 heads? 1 head? 2 heads? 3 heads? 4 heads?
Our possibilities are:
0 heads:
TTTT
1 head:
THHH
HTHH
HHTH
HHHT
2 heads:
TTHH
THTH
THHT
HTTH
3 heads:
THHH
HTHH
HHTH
HHHT
4 heads:
HHHH
HTHT
HHTT
Binomial Distribution
The binomial distribution is the probability distribution for the number of
successes in a sequence of Bernoulli trials.
Suppose that n Bernoulli trials are to be performed. Then the
number of outcomes with exactly x successes is equal to the
binomial coefficient:
n
 
 x
Therefore, the probability of getting x successes out of n trials is:
X
n x
B(n, p)  P( X  x)    p (1  p) n  x , x  0, 1, 2,..., n
 x
Given a binomial distribution with n = 11 and p = 0.4, find the
probability of getting exactly four successes.
Answer: 0.236
A fair die is rolled ten times. Find the probability that exactly three 6s
are scored.
Answer: 0.155
Ninety percent of the graduates of State University who apply to a
particular medical school are admitted. This year six graduates
have applied for admission to the medical school. Find the
probability that only four of them will be accepted.
Answer: 0.0984
A library contains 2000 fiction books and 3000 non-fiction books.
If eleven books are picked at random, estimate the probability that
four will be fiction.
Answer: 0.236
Mr. Galaty is taking a multiple choice exam that consists of five
questions. Each question has four possible answers. He guesses
at every answer. What is the probability that he passes the exam if
he needs at least four correct answers to pass?
Answer:
0.0156
In a school there are 600 girls and 500 boys. If ten are chosen
at random to go on a trip, estimate the probability that seven of
them will be girls.
Answer:
0.162
When Mary and Jane play tennis, the probability that Mary wins a point
is 0.4. Find the probability that Mary will win fewer than three of the
first twelve points.
Answer:
0.083
The New Progressive Party is supported by 5% of the population.
Find the probability that a random sample of 30 people contains at
most two supporters of the party.
Answer:
0.812
A coin is flipped three times. What are the possible probabilities of
getting a heads?
To find the mean, or expectation, we use:
x
0
1
2
3
P(x)
1
8
3
8
3
8
1
8
E( X )     x  P( X  x)
Therefore:
1
3
3
1
E ( X )  0   1  2   3 
8
8
8
8
3
Notice that:
E( X ) 
2
1
E( X )  3
2
Therefore, the formula for the mean of a binomial distribution is:
E ( X )  np
The variance of a discrete random variable, x, is defined by:
Var ( X )   x2 P( X  x)  2
2
3
3
1 3
 1
Var ( X )   0   1  4   9     
8
8
8 2
 8
 3 12 9   9   24   9  3
Var ( X )               
8 8 8 4  8  4 4
3
1 1
Var ( X )   3  
4
2 2
Therefore, the formula for the variance of a binomial
distribution is defined as:
Var ( X )  npq  np(1  p)
Therefore, the formula for the standard deviation of a binomial
distribution is defined as:
Var ( X )  npq  np(1  p)
The probability that an apple, picked at random from a sack, is bad is
0.05. Find the standard deviation of the number of bad apples in a
sample of 15 apples.
Answer: Standard Deviation = 0.844
In a group of people the expected number who wear glasses is 2 and the
variance is 1.6. Find the probability that 6 people in the group wear
glasses.
Answer: 0.00551
According to the US National Center for Health Statistic, about
60% of all eye operations are performed on females. Suppose
that three patients are to be selected at random. Let x denote the
number of patients out of the three chosen that are female.
(a) Find the mean of the random variable x.
(b) Find the standard deviation of the random variable x.
Answer: Mean = 1.8
Standard Deviation = 0.849
When a boy plays a game at a fair, the probability that he wins a
prize is 0.25. He plays the game 10 times. Let X denote the total
number of prizes that he wins. Assuming that the games are
independent, find:
(a)
E(X)
(b)
P ( X  2)
Answer:
(a) 2.5
(b) 0.526
M03/HL1/6
Marian shoots ten arrows at a target. Each arrow has probability 0.4
of hitting the target, independently of all other arrows. Let X denote
the number of these arrows hitting the target.
(a) Find the mean and standard deviation of X.
(b) Find
P ( X  2).
Answer:
(a) Mean = 4, Stan. Dev. = 1.55
(b) 0.954
M04/HL1/12