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Math 166 Lecture Notes 2.
Section 2.4 The Binomial Distribution This section of material brings us to a discussion of a certain class of experiment. When this type of experiment is involved, certain probabilities can be computed using special formulas and calculator functions. Caution: Not all experiments fall into this category. But, for those that do, the calculations learned in this section are much simpler than the calculations learned in earlier sections of material.
Bernoulli Trials In general, experiments with two outcomes are called Bernoulli trials. A sequence of Bernoulli (binomial) trials is called a binomial experiment.
A binomial experiment has the following properties:
- The number of trials in the experiment is fixed.
- There are two outcomes of the experiment: “success” and “failure.”
- The probability of success in each trial is the same.
- The trials are independent of each other.
Probabilities in Bernoulli Trials Let X be the variable that gives the number of successes in a binomial experiment. In this case, the variable X is called a binomial random variable, and the probability distribution of X is called a binomial distribution.
Let p = probability of success in any trial, and q = the probability of failure. Note: q = 1–p.
To find the probability of exactly x successes in n independent trials,
we compute: P ( X= x) =C(n,x)px^ qn−x (x = 0, 1, 2, …, n).
Example 1: A fair four-colored spinner (red, blue, green, yellow) is spun five times. Compute the probability of obtaining exactly two blues in the five spins.
Solution 1: Construct a tree diagram or list the events in the sample space. (too big!)
Solution 2: Compute the probability using counting methods and P(E)= n(E)/n(S):
C ⋅ ⋅ ⋅
Solution 3: Treat it as a binomial experiment, using P (X= x) =C(n,x)px^ qn−x
Note: The number of trials is fixed. (five spins) There are two outcomes. (blue is success, not blue is failure) The probability of success on each spin is the same. The trials are independent of each other (the results of one spin doesn’t affect the next)
We need to know the values of n, x, p, and q for this problem.
n = 5 trials (spins), x = 2 successes (2 blues), p = 4
(1 out of 4 sections is blue), q = 4
So, P ( X= x) =C(n,x)px^ qn−x= C(5, 2) x (.25) 2 x (.75) 3 = 10 x .0625 x .421875 ≈.
Solution 4: Again treating it as a binomial experiment, we will learn how to use a calculator function that performs the formula’s computations for us.
P( X= 2 ) =binom pdf (n, p, x) = binom pdf (5, .25, 2) ≈. 2637
Example 2: A fair five-sectioned spinner with numbers 1 through 5 is spun seven times. Find the probability of getting an odd number at least five times.
Solution 1: Draw a tree diagram, write out the entire sample space, and count!
Solution 2: Compute the probability using counting methods and P(E)= n(E)/n(S):
C ( , )⋅ ( ⋅ ⋅ ⋅ ⋅ )⋅ ⋅ +C( , )⋅( ⋅ ⋅ ⋅ ⋅ ⋅ )⋅ +C( , ) ⋅
Are there other options? Can we solve this as a binomial experiment? Note: The number of trials is fixed. (seven spins) There are two outcomes. (odd is success, even is failure) The probability of success on each spin is the same. The spins are independent of each other.
We want the probability of obtaining exactly 5 or exactly 6 or exactly 7 odd numbers.
What is X? What is n? What is p? What is x? What is q?
In mathematical symbols, we want P(X=5 or X=6 or X=7) or P(X=5) + P(X=6) + P(X=7)
Solution 3: Remember: P X( = x ) =C n x p q( , ) x^ n^ −x
We would need to work this out for each value, then add
P( X= 5 )=C( 7 , 5 )(. 6 )^5 (. 4 )^2 = 21 x .07776 x .16 =. 2612736
P( X= 6 )=C( 7 , 6 )(. 6 )^6 (. 4 )^1 = 7 x .046656 x .4 =. 1306368
P( X= 7 )=C( 7 , 7 )(. 6 )^7 (. 4 )^0 = 1 x .0279936 x 1 =. 0279936
. 4199040 ≈. 4199
Solution 4: There is also a calculator function that can give you the values above.
Use the 2nd^ function button and find the DISTR key, option 0: binompdf
P( X= 5 ) =binom pdf (7, .6, 5) =. 2612736
P( X= 6 ) =binom pdf (7, .6, 6) =. 1306368
P( X= 7 ) =binom pdf (7, .6, 7) =. 0279936
. 4199040 ≈. 4199
Solution 5: There is another calculator function that can help you in this type problem.
Use the 2nd^ function button and find the DISTR key, option A: binomcdf
P( X = 5 or 6 or 7 ) = 1 – P(X = 0 or 1 or 2 or 3 or 4) = 1 – binom cdf (7, .6, 4) = 1 –. 580096 =. 419904 ≈. 4199
Example 5: Of every 100 clothes dryers built at a particular factory, 78 of them will not have problems that require repair until after the 1 year manufacturer’s warranty is past. Cindy recently purchased 20 clothes dryers from that factory for her laundromat.
What is the probability that 1 year from now… a. None of Cindy’s machines will have needed repair? b. Exactly five of Cindy’s machines have needed repair? c. At least seventeen of them will not have needed repair? d. At least two of Cindy’s machines will have needed repair?
Solution: Is this a binomial distribution?
If so, what is X? What is n? What is x? What is p? What is q?
We should note that it is often helpful, especially in the more challenging problems, to create a quick table correlating the #’s of failures and the #’s of successes.
#bad 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 #good 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0
a.
b.
c.
d.
It is worth noting that each of the above problems could be worked with success and failure defined exactly the opposite of the ways worked above.
a.
b.
c.
d.
