Activity 6 - 8

How Many Boys (or Girls)?

The five children of Tsar Nicholas II, circa 1907; Romanov Collection, General Collection, Beinecke Rare Book and Manuscript Library, Yale University

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?

Now if the births are independent, then what if the probability that if a family had 2 kids, both are boys?

Fill in the following table for 2 children:

X (# of boys) P(x) Ways it happens

0

1

2

P(child = Boy) = 0.5

P(1st child is boy and 2nd child is boy) = 0.5 0.5 = 0.25

0.25 G G0.50 B G G B0.25 B B

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) Ways it happens

0

1

2

3

4

0.0675 gggg

0.0675 bbbb

0.25 gggb ggbg gbgg bggg

0.25 bbbg bbgb bgbb gbbb

0.375 bbgg bggb gbbg ggbb gbgb bgbg

English Phrases

Math Symbol

English Phrases

≥ At least No less than Greater than or equal to> More than Greater than< Fewer than Less than≤ No more than At most Less than or equal to= Exactly Equals Is ≠ Different from

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, x = 0, 1, 2, 3, …, n

nCx 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 n! = --------------k 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

Cumulative probability or cdf P(x ≤ A)

Complement Rule:cdf(x > A) = 1 – P(x ≤ A)

Values of Discrete Variable, X

P(X)

∑P(x) = 1

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 2aIn 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:

 

a) 6 will prefer Pepsi?

  

  

P(d=P) = 0.5

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 2bIn 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:

 

b) 12 will prefer Coke?

  

P(d=P) = 0.5

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

c) at least 15 will prefer Pepsi?

  

P(d=P) = 0.5 P(x) = nCx px(1-p)n-x

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

d) at most 8 will prefer Coke?

P(d=P) = 0.5 P(x) = nCx px(1-p)n-x

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 3aA 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: 

a) all 15 have the disease?  P(Y) = 0.9

P(x) = nCx px(1-p)n-x

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 3bA 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: 

   b) at least 13 people have the disease?

  

P(Y) = 0.9P(x) = nCx px(1-p)n-x

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

c) 8 have the disease?

P(Y) = 0.9 P(x) = nCx px(1-p)n-x

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.

(a)Find the probability that exactly four of the kangaroo’s young will survive to maturity

p = 0.2 n = 10 x = 4

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.

(b) Find the probability that at least four of the kangaroo’s young will survive to maturity.

p = 0.2 n = 10 x ≥ 4

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