Geometric Distribution

GEOMETRIC DISTRIBUTION

 

1. Geometric distribution

Suppose that independent trials, each having a probability p, 0 < p < 1, of being a success, are performed until a success occurs. If we get X equal to the number of trials required, then

P{X = x} = (1 − p)^{x - 1}p = q^{x – 1}p , x = 1,2,...

Equation follows because in order for X to equal n, it is necessary and sufficient that the first x - 1 trials are failures and the nth trial is a success. Equation then follows, since the outcomes of the successive trials are assumed to be independent,

Since

it follows that with probability 1, a success will eventually occur. Any xod & de viidedow random variable X whose probability mass function is given by equation is said to be a geometric random variable with parameter p.

Note: Geometric distribution has a important application in queueing theory, related to the number of units which are being served or waiting to be served at any given time.

 

2. Recurrence formula for Geometric Distribution.

We have, P(X = x) = pq^{x – 1}, p + q = 1, x = 1,2,....

P(X = x + 1) = pq^x

So P (X = x + 1)/ P(X = x) = q

P(X = x + 1) = q P (X = x)

If we know P(X = 0 )  then we can write down P (1), P (2) and so on.

Note 1: The sum of geometric random variables is negative Binomial, whereas sum of negative Binomial random variables is negative Binomial.

Note 2: Geometric distribution has memoryless property that if an event has not occurred during first r repetitions of an experiment E, then the probability that it will not occur during the next t repetitions is same as the probability that it will not occur during the first t repetitions.

 

Example 1.9.1

If X is a geometric variate taking values 1, 2, ... ∞, then find P(X is odd). [A.U A/M 2017 R-13]

Solution: We know that, for a geometric distribution P(X = r)q^{r - 1}p

P(X is odd) = P(X =1, 3, 5, ...)

= P(X = 1) + P(X = 3) + P(X = 5) +.....

= p + q²p + q⁴p +.....

== p (1+q² + q⁴ + ...)

=  p (1 - q²)^-1  = p /(1 + q)(1 - q) = 1/1 + q

 

Example 1.9.2

If the probability that a target is destroyed on any one shot is 0.5, what is the probability that it would be destroyed on 6^th attempt? [A.U N/D 2013] [A.U N/D 2017 R-8] [R13 RP]

Solution :

Given: p = 0.5; q = 10.5 = 0.5

P(X = r) = q^{r - 1} p

P(X = 6)  = q⁵ p = (1/2)⁵ ½ = 1/2⁶  = 1/64

 

Example 1.9.3

If the probability is 0.05 that a certain kind measuring device will show excessive drift, what is the probability that the sixth of these measuring devices tested will be the first to show excessive drift?

Solution : P = 0.05, q = 1- 0.05 = 0.95

P (X = r) = q^{r – 1}p

P(X = 6) = (0.95)⁵ (0.05)

 

Example 1.9.4.

Suppose that a trainee soldier shoots a target in an independent fashion. If the probability that the target is shot on any one shot is 0.8. (i) What is the probability that the target would be hit on 6th attempt ? (ii) What is the probability that it takes him less than 5 shots? (iii) What is the probability that it takes him an even number of shots ?  [A.U N/D 2006] [A.U CBT A/M 2011]

Solution :

p = 0.8; q = 1-p = 1 - 0.8 = 0.2

 P(X = r) = q ^{r -1}. p

 (i) The probability that the target would be hit on the 6^th attempt

P(X = 6) = (0.2)⁵ (0.8)

(ii) The probability that it takes him less than 5 shots = P ( X < 5 )

= (0.8) + (0.2) (0.8) + (0.2)² (0.8) + (0.2)³ (0.8)

= 0.9984

(iii) The probability that it takes him an even number of shots is

= P(X = 2) + P(X = 4) + P(X = 6) + ....

= (0.2) (0.8) [1 + (0.2)² + (0.2)⁴+...]

=(0.2) (0.8) [1 + (0.04) + (0.04)² + ....]

= (0.2) (0.8) [1-0.04] ̄^-1

= (0.2) (0.8) [0.96]^-1= 0.1667.

 

Example 1.9.5.

A and B shoot independently until each has hit his own target. The probabilities of their hitting the target at each shot 3/5 and 5/7 respectively. Find the probability that B will require more shots than A.

Solution: X → number of trials required by A to get his first success. Then X follows a geometric distribution given by,

Let Y→ number of trials required by B to get his first success. Then Y follows a geometric distribution given by,

P(B requires more trials to get his first success than A requires to get his first success)

 

Example 1.9.6

Establish the memoryless property of geometric distribution. [A.U Trichy A/M 2010] [A.U CH A/M 2011] [A.U N/D 2015 R13 RP] [A.U N/D 2018 R13 RP] [A.U M/J 2006] [A.U N/D 2010]

Sol. If X has a geometric distribution, then for any two positive integers 'm' and 'n', P[X > m + n /X > m] = p[X >n]

Proof :

P[X > m + n/ X > m] =

P[X > m + n / X > m] = q^{m+ n} /q^m = qⁿ

P[X > n] = qⁿ

Hence, P [X > m + n/ X > m] = P[X > n]

 

Example 1.9.7

A coin is tossed until the 1^st head occurs. Assuming that the tosses are independent and the probability of a head occuring is p, find the value of p so that the probability that an odd number of tosses is required, is equal to 0.6. Can you find a value of p, so that the probability is 0.5 that an odd number of tosses is required? [AU May 2004, N/D 2010]

Solution: Let X denote the number of tosses required to get the first head X follows geometric distribution.

P[X = x] =  q^{x - 1}p, x = 1, 2, ..., & p + q = 1 ....(1)

PIX = an odd number of tosses]

= P[X= 1] + P[X = 3] + P[X = 5] + ...

= p + q² p + q⁴ p + ..

= p[1 + q² + q⁴ + ...]

= P[1- q ²]^-1  [ (1- x)^-1 = 1 + x + x² + ...]

= p/1 – q²

= p/(1 +  q) (1 - q) = 1/1 + q  [p + q = 1]

Given : P[X = odd number of tosses] = 0.6

⇒ 1/ 1 + q = 0.6 = 3/5

⇒ 1 =  (3/5)(1 + q)

⇒ 5/3 – 1 = q

⇒ 2/3 = q

P = 1 – q = 1 – 2/3 = 1/3

To find p if P[X = odd number of tosses] = 0.5

(2) ⇒ 1/ 1 + q = 0.5 = 1/2

=> 1 = (1/2)(1 + q)

⇒ 2 = 1 + q

=> q = 1⇒ p = 0

which is meaningless. So value of p cannot be found out.

 

Example 1.9.8

If X₁, X₂ be independent random variables each having geometric distribution q^Kp, K = 0, 1, 2, Show that the conditional distribution of X₁ given X₁+ X₂ is uniform. [A.U N/D 2006]

Solution: Given: P(X₁ = K) = P(X₂ = K) = pq_K, K = 0,1,2,...

(i.e.,) P(X₁ = K) = P(X₂ = K) = pq^{K - 1}, K = 1, 2, ...

X₁+ X₂ = n is a discrete uniform distribution.

Note: When P[X =r] = constant, the discrete random variable X is said to follow a discrete uniform distribution.

Thus the conditional distribution of X, given that X + Y = n, is a discrete uniform distribution.

 

Example 1.9.9

If the probability that an applicant for a driver's licence will pass the road test on any given trial is 0.8, what is the probability that he will finally pass the test (a) on the fourth trial and (b) in fewer than 4 trials ? [A.U Trichy A/M 2010] [A.U A/M 2010, N/D 2012] [A.U A/M 2015 (RP) R13] [A.U A/M 2017 R-08]

Solution: Let X denote the number of trials required to achieve the first success. Then X is a geometric distribution given by

P (X = r) = q^{r - 1}p; r = 1, 2, 3, ..., ∞

Here p = 0.8 and q = 0.2

(a) P (X = 4) =(0.2)^{4 - 1} (0.8) (0.2)³ (0.8) = 0.0064

(b) P (X < 4) = P[X = 1] + P(X = 2] + P [X = 3]

= (0.2)^{1 - 1}0.8+ (0.2)^{2 - 1} (0.8) + (0.2)^{3 - 1} (0.8)

= [1 +0.2 + (0.2)²] (0.8) =  0.992