Transcript P(child = Boy)

Activity 6 - 8
The five children of Tsar Nicholas II, circa 1907; Romanov Collection, General
Collection, Beinecke Rare Book and Manuscript Library, Yale University
How Many Boys (or Girls)?
Objectives
• Recognize components of a binomial experiment
• Calculate binomial probabilities
Vocabulary
• Binomial Probability Distribution – a discrete (integer
only) probability distribution that meets four specific
criteria
• Binomial Coefficient – the number of different ways a
certain combination can be done. nCr is n things
taken r at a time.
• Independent Events – when the outcome of one
event has no effect on the outcome of another event
Activity
In a family with 4 children, how unusual would it be for all
children to be boys? Or by the same token, all girls?
Such questions are relevant for many families in cultures
around the world, especially when one gender is favored
over another. (Note: China’s cultural preference for male
children has led to huge imbalance between young males
and females in China)
Suppose that whether a baby is a boy or a girl is equally
likely.
What is the probability that a single child is a boy?
P(child = Boy) = 0.5
Activity cont
What is the probability of the second child born being a
boy?
P(child = Boy) = 0.5
Now if the births are independent, then what if the
probability that if a family had 2 kids, both are boys?
P(1st child is boy and 2nd child is boy) = 0.5  0.5 = 0.25
Fill in the following table for 2 children:
X (# of boys)
P(x)
0
0.25
0.50
0.25
1
2
Ways it happens
GG
BG
BB
GB
Activity cont
Back to the original question about families with 4 kids.
Fill in the following table for 4 children:
X (# of boys)
P(x)
0
0.0675
0.25
0.375
0.25
0.0675
1
2
3
4
Ways it happens
gggg
gggb ggbg gbgg bggg
bbgg bggb gbbg ggbb gbgb bgbg
bbbg bbgb bgbb gbbb
bbbb
English Phrases
Math
Symbol
≥
At least
>
More than
<
Fewer than
≤
No more than
=
Exactly
≠
Different from
English Phrases
No less than
Greater than
Less than
At most
Equals
Greater than or equal to
Less than or equal to
Is
Example 1
Translate into mathematical symbols:
A. At least 5 quarters
B. No more than 3 nickels
C. fewer than 6 pennies
D. Something different than 2 dimes
$ ≥ 5 quarters
$ ≤ 3 nickels
$ < 6 pennies
$ ≠ 2 dimes
Binomial Probability Criteria
An experiment is said to be a binomial experiment
provided:
1. The experiment is performed a fixed number of
times. Each repetition is called a trial.
2. The trials are independent
3. For each trial there are two mutually exclusive
(disjoint) outcomes: success or failure
4. The probability of success is the same for each trial
of the experiment
Binomial PDF
The probability of obtaining x successes in n
independent trials of a binomial experiment, where the
probability of success is p, is given by:
P(x) = nCx px (1 – p)n-x,
nCx
x = 0, 1, 2, 3, …, n
is also called a binomial coefficient and is defined by
combination of n items taken x at a time or
where n! is n  (n-1)  (n-2)  …  2  1
n
k
n!
= -------------k! (n – k)!
TI-83 Binomial Support
• For P(X = k) using the calculator: 2nd VARS
binompdf(n,p,k)
• For P(k ≤ X) using the calculator: 2nd VARS
binomcdf(n,p,k)
• For P(X ≥ k) use 1 – P(k < X) = 1 – P(k-1 ≤ X)
TI-83 Binomial Functions
• Cumulative probability density function (cdf)
–
–
–
–
sum of the probability for values less than x
2nd VARS (DISTR), arrow down to A: binomcdf
parameters: # of trials, p(s), x
probability of 4 or less heads in 6 flips of a coin:
binomcdf(6,0.5,4)
• Probability density function (pdf)
–
–
–
–
probability for an “x =“ problem
2nd VARS (DISTR), arrow down to 0: binompdf
parameters: # of trials, p(s), x
probability of 4 heads in 6 flips of a coin:
binompdf(6,0.5,4)
Binomial Probabilities
∑P(x) = 1
P(X)
Cumulative
probability
or cdf
P(x ≤ A)
Values of Discrete Variable, X
Complement Rule:
cdf(x > A) = 1 – P(x ≤ A)
X=A
• Binomial numbers are discrete (integers only)
• P(x ≥ 4) = 1 – P(x ≤ 4)
• P(x < 4) = P(0) + P(1) + P(2) + P(3)
• Need to use our calculator!
Example 2a
In the “Pepsi Challenge” a random sample of 20 subjects
are asked to try two unmarked cups of pop (Pepsi and
Coke) and choose which one they prefer. If preference is
based solely on chance what is the probability that:
P(d=P) = 0.5
a) 6 will prefer Pepsi?
P(x) = nCx px(1-p)n-x
P(x=6 [p=0.5, n=20]) = 20C6 (0.5)6(1- 0.5)20-6
= 20C6 (0.5)6(0.5)14 = 0.037
Example 2b
In the “Pepsi Challenge” a random sample of 20 subjects
are asked to try two unmarked cups of pop (Pepsi and
Coke) and choose which one they prefer. If preference is
based solely on chance what is the probability that:
P(d=P) = 0.5
b) 12 will prefer Coke?
P(x) = nCx px(1-p)n-x
P(x=12 [p=0.5, n=20]) = 20C12 (0.5)12(1- 0.5)20-12
= 20C12 (0.5)12(0.5)8 = 0.1201
Example 2c
P(d=P) = 0.5
P(x) = nCx px(1-p)n-x
c) at least 15 will prefer Pepsi?
P(at least 15) = P(15) + P(16) + P(17) + P(18) + P(19) + P(20)
Use cumulative PDF function in calculator
P(X ≥ 15) = 1 – P(X ≤ 14) = 1 – 0.9793 = 0.0207
Example 2d
P(d=P) = 0.5
P(x) = nCx px(1-p)n-x
d) at most 8 will prefer Coke?
P(at most 8) = P(0) + P(1) + P(2) + … + P(6) + P(7) + P(8)
Use cumulative PDF function in calculator
P(X ≤ 8) = 0.2517
Example 3a
A certain medical test is known to detect 90% of the people
who are afflicted with disease Y. If 15 people with the
disease are administered the test what is the probability
that the test will show that:
P(x) = nCx px(1-p)n-x
P(Y) = 0.9
a) all 15 have the disease?
P(x=15 [p=0.9, n=15]) = 15C15 (0.9)15(1- 0.9)15-15
= 15C15 (0.9)15(0.1)0 = 0.20589
Example 3b
A certain medical test is known to detect 90% of the people
who are afflicted with disease Y. If 15 people with the
disease are administered the test what is the probability
that the test will show that:
P(x) = nCx px(1-p)n-x
P(Y) = 0.9
b) at least 13 people have the disease?
P(at least 13) = P(13) + P(14) + P(15)
Use cumulative PDF function in calculator
P(X ≥ 13) = 1 – P(X ≤ 12) = 1 – 0.1841 = 0.8159
Example 3c
P(Y) = 0.9
P(x) = nCx px(1-p)n-x
c) 8 have the disease?
P(x=8 [p=0.9, n=15]) = 15C8 (0.9)8(1- 0.9)15-8
= 15C8 (0.9)8(0.1)7 = 0.000277
Example 4a
Suppose that in its lifetime, a female kangaroo gives
birth to exactly 10 young. Suppose further that each
kangaroo baby, independently of all the others, has a
20% chance of surviving to maturity.
p = 0.2 n = 10 x = 4
(a) Find the probability that exactly four of the
kangaroo’s young will survive to maturity
With x = 4, then we must use binompdf,
binompdf(10, 0.2, 4) = 0.0881
Example 4b
Suppose that in its lifetime, a female kangaroo gives
birth to exactly 10 young. Suppose further that each
kangaroo baby, independently of all the others, has a
20% chance of surviving to maturity.
p = 0.2 n = 10 x ≥ 4
(b) Find the probability that at least four of the
kangaroo’s young will survive to maturity.
With x ≥ 4, then we must use binomcdf and the
complement rule,
binomcdf(10, 0.2, 3) = 0.8791
P(x ≥ 4) = 1 – P(x ≤ 3) = 1 – 0.8791 = 0.1209
Example 5
State the conditions that must be satisfied for a
random variable X to have a binomial distribution.
1) Outcomes are mutually exclusive (success or failure)
2) Probability of success is the same for each event
3) Each event is independent
4) Fixed number of trials
Summary and Homework
• Summary
– If two independent events, then P(A and B)= P(A)P(B)
– Binomial Conditions:
•
•
•
•
Fixed number of trials, n
Each trial independent
Each trial has two mutually exclusive outcomes (success/fail)
Probability of success, p, is constant (P(failure) = (1 – p) = q
– Binomial Coefficient is the combination of n things
taken r at a time
– Binomial probability; P(x) = nCx pnqn-x
• Homework
– pg 769-771; problems 1-4, 7