Lecture 11- The Binomial Distribution

Transcription

1 Lecture - The Binomial Distribution Now that we have seen the basic ideas behind a discrete probability distribution we are going to see our first example: the binomial distribution. The binomial distribution is a probability distribution of a discrete random variable. We will use the random variable to count the number of successes out of so many trials. Definition- A binomial experiment has the following characteristics.. There are a fixed number of trials which we denote by n.. Each trial is independent (in the sense of probability) of all the other trials. 3. Each trial has only two possible outcomes. We call one outcome success and denote its probability p. We call the other outcome failure and denote its probability q. 4. For each trial the probability of success and failure are the same. In addition we must have p+ q = and q = p. We flip a fair coin 00 times and want to know the probability of getting 60 or more heads. This is a binomial experiment.. Here we have a fixed number of trials, n = 00.. Each time we flip the coin the outcome is independent from any other flip. 3. Each trial has two outcomes; heads or tails. In our experiment flipping a head is a success and flipping a tail is a failure. 4. The probabilities of success and failure are the same for each trial. Here and q =. We also see that p+ q = and q = p. p = For any binomial experiment we always ask what is the probability of r successes out of n trials. Here is an example. Let s say we flip a fair coin 3 times and we want to know the probability of getting or more heads.. st we have n = 3 for 3 flips.. Now we wish to make or more heads. How can we do this? HHH, HHT, HTH, THH will all work. 3. Now we must ask what is the probability of each of these combinations of flips? PHHH = PHPHPH = = P( HHT ) = PHTH = PTHH =

2 4. We asked what is the probability of getting or more heads. Said another way P( HHH or HHT or HTH or THH ) = P( HHH ) + P( HHT ) + P( HTH ) + P( THH ) = Here is an easier way to do this. Theorem- Let s say that we have a binomial experiment where p is the probability of a success and q is the probability of a failure. If we have n trials and want to know Pr (the probability of exactly r successes) then r n r Pr () = ncr p q Where n n! ncr = = r ( n r)! r! Consider the following experiment. We have a group of 6 people and want to know what the chance of randomly selecting 3 people with brown eyes is. p = probability of having brown eyes =.6 q = probability of not having brown eyes = -.6 =.4 So we consider picking someone with brown eyes to be a success. Thus we are trying to find P(3). Here we have n = 6 and r = 3 so ! 3 3 P( 3 ) = 6C3(.6) (.4) = (.6) (.4) = !3! Note- This formula only works for finding the probability of exactly r successes. It will not work for finding the probability of at most or at least r successes. You can then ask the question how do we handle each of these cases. This is easy since the nature of a binomial experiment says that its events are mutually exclusive. Let s say that n =0 and we want to find ( 5) = ( 5) + ( 6) + ( 7) + + ( 9) + ( 0) ( 4) = + + ( 3) + ( 4) P r P P P P P P P r P P P P In general we will have two cases. Theorem- Given a binomial experiment with n trials and we want to calculate the probability of a range of successes we have one of the following.. At least r successes, P( r number of successes) then P( r number of successes) = P( r) + P( r + ) + + P( n)

3 . At most successes, P r number of successes r + + ( P r number of successes = P + P P r) A survey reports that 5% of Americans are afraid of being alone in the house at night. If a random sample of 0 Americans is selected find the probability that There are at most 3 people in the sample who are afraid of being home alone at night. There are at least 3 people in the sample who are afraid of being home alone at night. Here n = 0, p =.05, and.95. The first part is asking for P r so we have q = ( 3) ( 3) = ( 0) ( 3) P r P P P P (.05) (.95) (.05) (.95) (.05) (.95) (.05) (.95) = C + C + C + C = =.94 For the second part we are asked to find P r 3 = P 3 + P 4 + P 5 + P P 0 This is way to long. Here is the shortcut. This problem can be solved by finding P( 0) + P + P and subtracting it from. P( 0) + P + P( ) = = =.076

4 Lecture Properties of the Binomial Distribution For any distribution we can talk about the mean, standard deviation, and variance. Let s do it for the binomial distribution. Due to the nature of the binomial distribution these calculations turn out to be very easy. Here are the formulas Here we have n = The number of trials p = Probability of success q = Probability of failure = -p μ = np x = np σ = npq s = npq σ = npq s = npq Examples- A person flips a weighted coin 6 times. So we have p = heads =.7 q = tails =.3 Since there are 6 flips we have 6 trials. We would like to know the mean, standard deviation, and variance for the distribution associated to this experiment. x= np= 6.7 = 4. s = npq = 6.7.3=. s =. =.5 Here is a more complicated example which requires us to use a little bit of algebra. A satellite requires three solar cells for its power. The probability that any one of these cells will fail is 0.5, and the cells operate or fail independently. What is the least number of cells the satellite should have so that the expected number of working cells is no smaller than three? The magic numbers is any binomial experiment are n, p, q, and r. Let s identify them. Here n is the number of cells, p is the probability that a cell will work, q is the probability that a cell will fail, and r is the number of working cells. Recall that the expected value is μ. We are asked to find a value for n so that μ is no smaller than three. Since μ = np we need only find n and p. st we know q = 0.5. Since p = qwe have q = 0.5 nd we want the expected value to be no smaller than three or μ 3. We use this to find the value for n. We must solve the following.

5 3 μ 3 np ( 5) 3 n 0. 3 n n So we have found that3.53 n. Since n represents the number of cells on the satellite it must be a whole number, so we round up. Thus n = 4.