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“Teach A Level Maths” Statistics 1 The Geometric Distribution © Christine Crisp The Geometric Distribution Statistics 1 OCR "Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used with permission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages" The Geometric Distribution When playing some board games we have to roll a six with a die before we can begin. 1 The probability of rolling a six with the 1 try is . 6 st The probability of not getting a six directly but getting one with the 2nd try is 5 1 6 6 and failing twice but succeeding the 3rd time is 2 and so on. 5 5 1 5 1 6 6 6 6 6 The Geometric Distribution In general, if X is the random variable “ the number of trials to success”, and if the probability of success is p and the probability of failure is q, where p + q = 1, then P ( X 1) p P ( X 2) qp 2 q P ( X 3) p and so on. This distribution is called the geometric distribution. The general term is given by P ( X x ) q x 1 p x 1 p Substituting for q, we get P ( X x ) (1 p) In your formulae book this is written as p(1 p) x 1 The next page proves that X is a random variable but if you haven’t studied the Geometric Series in Pure Maths, skip over it. SKIP The Geometric Distribution The sequence of probabilities for X = 1, 2, 3, . . . form an infinite geometric sequence: p , qp , q 2 p , . . . The sum of an infinite number of terms is a S 1 r where a is the 1st term and r is the common difference. So, since a p and r q p S 1 q Since q = 1 p, we have 1 – q = p, so, S 1 The sum of the probabilities equals 1, the condition for a random variable. The Geometric Distribution The Mean and Variance of the Geometric Distribution The mean and variance of the Geometric Distribution are given by mean, 1 p variance, 2 1 p p2 A random variable, X, with a geometric distribution is written as X ~ Geo( p) The Geometric Distribution Finding P ( X x ) There are 2 ways of reasoning to get to the result. 1 e.g. Suppose we have X ~ Geo and we want 6 P ( X 5) Method 1: P ( X 5) 1 P ( X 4) 1 ( p qp q 2 p q 3 p) 1 5 p q so, using a calculator, we get 6 6 P ( X 5) 1 0 518 0 482 ( 3 s. f . ) Method 2: The probability of success on the 5th trial, or the 6th, or the 7th etc. means we have failure on the 1st 4, so, P ( X 5) q 4. We get the same answer. The Geometric Distribution SUMMARY If X is a random variable “ the number trials to success” then P ( X 1) p P ( X 2) qp 2 q P ( X 3) p and so on. where, p is the probability of success and q = 1 – p is the probability of failure. We write X ~ Geo( p) 1 The mean of X is given by p 1 p 2 The variance of X is given by p2 ( These results are in the formulae booklet ) Also, P ( X x ) q x 1 The Geometric Distribution Exercise 1. When a darts player aims at a “double 20” on the board the probability of hitting it is 0·2. (a) What is the probability of hitting the double 20 (i) for the 1st time with the 3rd dart, (ii) not hitting it in 10 tries? (b) Find the expected number of throws before the double 20 is hit. Solution: Let X be the random variable “ number of times until the double 20 is hit” Then, X ~ Geo (0 2) (a)(i) P ( X 3) q 2 p (0 8) 2 (0 2) 0 128 The Geometric Distribution (a) What is the probability of hitting the double 20 (i) for the 1st time with the 3rd dart, (ii) not hitting it in 10 tries? (b) Find the expected number of throws before the double 20 is hit. X ~ Geo (0 2) (a)(ii) We need P ( X 11) q 10 0 8 10 0 107 1 (b) E ( X ) p 5 The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied. The Geometric Distribution In general, if X is the random variable “ the number of trials to success”, and if the probability of success is p and the probability of failure is q, where p + q = 1, then P ( X 1) p P ( X 2) qp 2 q P ( X 3) p and so on. This distribution is called the geometric distribution. The general term is given by P ( X x ) q x 1 p x 1 ( 1 p ) p Substituting for q, we get P ( X x ) In your formula book this is written as p(1 p) x 1 The Geometric Distribution Finding P ( X x ) There are 2 ways of reasoning to get to the result. 1 e.g. Suppose we have X ~ Geo and we want 6 P ( X 5) Method 1: P ( X 5) 1 P ( X 4) 1 ( p qp q 2 p q 3 p) 1 5 p q so, using a calculator, we get 6 6 P ( X 5) 1 0 518 0 482 ( 3 s. f . ) Method 2: The probability of success on the 5th trial, or the 6th, or the 7th etc. means we have failure on the 1st 4, so, P ( X 5) q 4. We get the same answer. The Geometric Distribution SUMMARY If X is a random variable “ the number trials to success” then P ( X 1) p P ( X 2) qp 2 q P ( X 3) p and so on. where, p is the probability of success and q = 1 – p is the probability of failure. We write X ~ Geo( p) 1 The mean of X is given by p 1 p 2 The variance of X is given by p2 ( These results are in the formula booklet ) Also, P ( X x ) q x 1