Probability Mass Function

5.1 The Bernoulli and Binomial Random Variables

Suppose that a trial, or an experiment, whose outcome can be classified as either a “success” or as a “failure” is performed. If we let X = 1 when the outcome is a success and X = 0 when it is a failure, then the probability mass function of X is given by

(5.1.1)P{X=0}=1−pP{X=1}=p

where p, 0 ≤ p ≤ 1, is the probability that the trial is a “success.”

A random variable X is said to be a Bernoulli random variable (after the Swiss mathematician James Bernoulli) if its probability mass function is given by Equations 5.1.1 for some p ∈ (0, 1). Its expected value is

E[X]=1·P{X=1}+0·P{X=0}=p

That is, the expectation of a Bernoulli random variable is the probability that the random variable equals 1.

Suppose now that n independent trials, each of which results in a “success” with probability p and in a “failure” with probability 1 – p, are to be performed. If X represents the number of successes that occur in the n trials, then X is said to be a binomial random variable with parameters (n, p).

The probability mass function of a binomial random variable with parameters n and p is given by

(5.1.2)P{X=i}=(ni) pi(1−p)n−i,i=0,1,…,n

where (ni)=n![i!(n−i)!] is the number of different groups of i objects that can be chosen from a set of n objects. The validity of Equation 5.1.2 may be verified by first noting that the probability of any particular sequence of the n outcomes containing i successes and n – i failures is, by the assumed independence of trials, pⁱ (1 – p)ⁿ⁻ⁱ. Equation 5.1.2 then follows since there are (ni) different sequences of the n outcomes leading to i successes and n – i failures — which can perhaps most easily be seen by noting that there are (ni) different selections of the i trials that result in successes. For instance, if n = 5, i = 2, then there are (52) choices of the two trials that are to result in successes — namely, any of the outcomes

(s,s,f,f,f)(f,s,s,f,f)(f,f,s,f,s)(s,f,s,f,f)(f,s,f,s,f)(s,f,f,s,f)(f,s,f,f,s)(f,f,f,s,s)(s,f,f,f,s)(f,f,s,s,f)

where the outcome (f, s, f, s, f) means, for instance, that the two successes appeared on trials 2 and 4. Since each of the (52) outcomes has probability p²(1 – p)³, we see that the probability of a total of 2 successes in 5 independent trials is (52)p2(1−p3). As a check, note that, by the binomial theorem, the probabilities sum to 1; that is,

∞∑i=0p(i)=n∑i=0(ni)pi(1−p)n−i=[p+(1−p)]n=1

The probability mass function of three binomial random variables with respective parameters (10, .5), (10, .3), and (10, .6) are presented in Figure 5.1. The first of these is symmetric about the value .5, whereas the second is somewhat weighted, or skewed, to lower values and the third to higher values.

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Figure 5.1. Binomial probability mass functions.

Example 5.1a

It is known that disks produced by a certain company will be defective with probability .01 independently of each other. The company sells the disks in packages of 10 and offers a money-back guarantee that at most 1 of the 10 disks is defective. What proportion of packages is returned? If someone buys three packages, what is the probability that exactly one of them will be returned?

Solution

If X is the number of defective disks in a package, then assuming that customers always take advantage of the guarantee, it follows that X is a binomial random variable with parameters (10, .01). Hence the probability that a package will have to be replaced is

P{X>1}=1−P{X=0}−P{X=1}=1−(100)(.01)0(.99)10−(101)(.01)1(.99)9≈.005

Because each package will, independently, have to be replaced with probability .005, it follows from the law of large numbers that in the long run .5 percent of the packages will have to be replaced.

It follows from the foregoing that the number of packages that will be returned by a buyer of three packages is a binomial random variable with parameters n = 3 and p = .005. Therefore, the probability that exactly one of the three packages will be returned is (31)(.005)(.995)2=.015.■

Example 5.1b

The color of one’s eyes is determined by a single pair of genes, with the gene for brown eyes being dominant over the one for blue eyes. This means that an individual having two blue-eyed genes will have blue eyes, while one having either two brown-eyed genes or one brown-eyed and one blue-eyed gene will have brown eyes. When two people mate, the resulting offspring receives one randomly chosen gene from each of its parents’ gene pair. If the eldest child of a pair of brown-eyed parents has blue eyes, what is the probability that exactly two of the four other children (none of whom is a twin) of this couple also have blue eyes?

Solution

To begin, note that since the eldest child has blue eyes, it follows that both parents must have one blue-eyed and one brown-eyed gene. (For if either had two brown-eyed genes, then each child would receive at least one brown-eyed gene and would thus have brown eyes.) The probability that an offspring of this couple will have blue eyes is equal to the probability that it receives the blue-eyed gene from both parents, which is (12)(12)=14. Hence, because each of the other four children will have blue eyes with probability 14, it follows that the probability that exactly two of them have this eye color is

(42)(1/4)2(3/4)2=27/128

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Example 5.1c

A communications system consists of n components, each of which will, independently, function with probability p. The total system will be able to operate effectively if at least one-half of its components function.

(a)

For what values of p is a 5-component system more likely to operate effectively than a 3-component system?

(b)

In general, when is a 2k + 1 component system better than a 2k – 1 component system?

Solution

(a)

Because the number of functioning components is a binomial random variable with parameters (n, p), it follows that the probability that a 5-component system will be effective is

(53)p3(1−p)2+(54)p4(1−p)+p5

whereas the corresponding probability for a 3-component system is

(32)p2(1−p)+p3

Hence, the 5-component system is better if

10p3(1−p)2+5p4(1−p)+p5≥3p2(1−p)+p3

which reduces to

3(p−1)2(2p−1)≥0

or

p≥12

(b)

In general, a system with 2k + 1 components will be better than one with 2k – 1 components if (and only if) p≥12. To prove this, consider a system of 2k + 1 components and let X denote the number of the first 2k – 1 that function. Then

P2k+1(effective)=P{X≥k+1}+P{X=k}(1−(1−p)2)+P{X=k−1}p2

which follows since the 2k + 1 component system will be effective if either

(1)

X ≥ k + 1;

(2)

X = k and at least one of the remaining 2 components function; or

(3)

X = k – 1 and both of the next 2 function.

Because

P2k−1(effective)=P{X≥k}=P{X=k}+P(X≥k+1)

we obtain that

P2k+1(effective)−P2k−1(effective)=P{X=k−1}p2−(1−p)2P{X=k}=(2k−1k−1)pk−1(1−p)kp2−(1−p)2(2k−1k)pk(1−p)k−1=(2k−1k)pk(1−p)k[p−(1−p)]since (2k−1k−1)=(2k−1k)≥0⇔p≥12

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Example 5.1d

Suppose that 10 percent of the chips produced by a computer hardware manufacturer are defective. If we order 100 such chips, will X, the number of defective ones we receive, be a binomial random variable?

Solution

The random variable X will be a binomial random variable with parameters (100, .1) if each chip has probability .9 of being functional and if the functioning of successive chips is independent. Whether this is a reasonable assumption when we know that 10 percent of the chips produced are defective depends on additional factors. For instance, suppose that all the chips produced on a given day are always either functional or defective (with 90 percent of the days resulting in functional chips). In this case, if we know that all of our 100 chips were manufactured on the same day, then X will not be a binomial random variable. This is so since the independence of successive chips is not valid. In fact, in this case, we would have

P{X=100}=.1P{X=0}=.9

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Since a binomial random variable X, with parameters n and p, represents the number of successes in n independent trials, each having success probability p, we can represent X as follow:

(5.1.3)X=n∑i=1Xi

where

Xi={1if the ith is a success0otherwise

Because the X_i, i = 1,…, n are independent Bernoulli random variables, we have that

E[Xi]=P{Xi=1}=pVar(Xi)=E[X2i]−p2=p(1−p)

where the last equality follows since X2i=Xi, and so E[X2i]=E[Xi]=p.

Using the representation Equation 5.1.3, it is now an easy matter to compute the mean and variance of X:

E[X]=n∑i=1E[Xi]=np

Var(X)=n∑i=1Var(Xi)since the Xi are independent=np(1−p)

If X₁ and X₂ are independent binomial random variables having respective parameters (n_i,p), i = 1, 2, then their sum is binomial with parameters (n₁ + n₂, p). This can most easily be seen by noting that because X_i, i = 1, 2, represents the number of successes in n_i independent trials each of which is a success with probability p, then X₁ + X₂ represents the number of successes in n₁ + n₂ independent trials each of which is a success with probability p. Therefore, X₁ + X₂ is binomial with parameters (n₁ + n₂, p).