I need help understaning how to solve these types of 
problems...can someone show me thanks... 

I can't tell whether you are to use binomial tables, 
formulas, TI graphical calculator, Excel, or normal tables 
to approximate the binomial probabilities.  So I'll just 
do them all by formula.

Problem #1
A coin is tossed 6 times. Find the probabilities of 
getting the following. 
a) Exactly 3 heads.

First we decide what is to be called a "success".
Here we will let a success be "getting a head".  
The probability of getting 1 success (head) in 1 
trial is .          

The probability of getting exactly x successes in 
n trials, where the probability of getting 1 success 
in 1 trial is p is given by this formula:



The probability of getting exactly 3 successes in 
6 trials, where the probability of getting 1 success 
in 1 trial is , so we substitute
x=3, n=6, and p= 














5/16


Problem #2
Quality Control A factory tests a random sample of 
20 transistors for defective transistors. The 
probability that a particular transistor will be 
defective has been established by past experience 
as .05.

a) What is the probability that there are no defective 
transistors in the sample? 

Another binomial problem.

First we decide what is to be called a "success".  
Here we will let a success be "getting a good 
(non-defective) transistor."  

We are not given the probability of 1 success in 1 
trial. Instead we are given the probability of 1 failure 
in 1 trial is ,05, so the probability of 1 success
in 1 trial is 1-.05 or .95 

The probability of getting exactly x successes in 
n trials, where the probability of getting 1 success 
in 1 trial is p is given by the formula as for the 
other problem:



The probability of getting exactly 20 successes in 
20 trials. So the probability of getting 1 success 
in 1 trial is , so we substitute
x=20, n=20, and p= 












(((.3584859224}}}

Problem#3
Flu Inoculations A flu vaccine has a probability of 80%
of preventing a person who is inoculated from getting the
flu. A county health office inoculates 83 people. Find the
probabilities of the following. 
a. Exactly 10 of the people inoculated get the flu.

First we decide what is to be called a "success".  Here 
we will let a success be "not getting the flu".  
The probability of getting 1 success in 1 trial is 80% 
or .          

As before, the probability of getting exactly x successes 
in n trials, where the probability of getting 1 success 
in 1 trial is p is given by this formula:



We are not being asked for the probability of exactly 
10 successes, but are asked for exactly 10 failures.  
So since there are 83 trials, we want the probability 
that there are 83 - 10 or 73 successes.

the probability of 1 success in 1 trial is .8.

The probability of getting exactly 73 successes in 83 
trials, where the probability of getting 1 success in 
1 trial is .8, so we substitute
x=73, n=83, and p=.8 











That comes out to .020984176

b. No more than 4 of the people inoculated get the flu.

This means we have either exactly 83 successes, exactly 
82 successes, exactly 81 successes, or exactly 80 successes.

To find P(83 or 82 or 91 or 80) we must add

P(83) + P(82) + P(81) + P(80)

I won't go through all the calculations, for they are the same:

For P(83), we use n = 83, x = 83, p = .8, that gives 9.046256972×10-9

For P(82), we use n = 83, x = 82, p = .8, that gives 1.877098322×10-7

For P(81), we use n = 83, x = 81, p = .8, that gives 1.92402578×10-6

For P(80), we use n = 83, x = 80, p = .8, that gives 1.29871717401×10-5

Adding those up gives 1.510795588×10-5

c. None of the people inoculated get the flu.

That was answered above, as that is the probability 
of 83 successes in 83 trials:

For P(83), we use n = 83, x = 83, p = .8, that gives 9.046256972×10-9

Edwin