Lesson 2: Binomial Distribution S2 – Chapter 2 – Binomial Distribution Lesson 2: Binomial Distribution

Frost True Stories Back in 2010 I was on holiday in Hawaii and visited the family of a friend. We noticed that at the dinner table that out of the 8 of us, 6 of us were left-handed (including myself). One of them asked, “The chances of that must be very low”. I saw that as a challenge. Consider the 8 people in a line. Suppose 10% of the population is left-handed. Can you now work out the probability of 6 being left-handed now? ? One possibility say is LLRLLRLL. The probability of this particular possibility is 𝟎.𝟏×𝟎.𝟏×𝟎.𝟗×𝟎.𝟏×…= 𝟎.𝟏 𝟔 × 𝟎.𝟗 𝟐 But there are 𝟖 𝟔 =𝟐𝟖 possible ways in which 6 people in the room could be left handed. So probability is 𝟖 𝟔 × 𝟎.𝟏 𝟔 × 𝟎.𝟗 𝟐 =𝟎.𝟎𝟎𝟎𝟐𝟐𝟔𝟖 ≈𝟏 𝒊𝒏 𝟒𝟒 𝟎𝟎𝟎

Test Your Understanding Q I throw a unfair coin 10 times, where the probability of heads is 𝑝. What’s the probability of throwing 3 heads? Let 𝑋 be a random variable which counts the number of heads seen in 10 throws. Determine the probability distribution. For convenience you can use 𝑞 as the probability of tails (where 𝑞=1−𝑝) 𝒙 𝟎 𝟏 𝟐 𝟑 𝟒 𝟓 𝟔 𝟕 𝟖 𝟗 𝟏𝟎 𝑃 𝑋=𝑥 𝑞 10 10 1 𝑝 𝑞 9 10 2 𝑝 2 𝑞 8 10 3 𝑝 3 𝑞 7 10 4 𝑝 4 𝑞 6 10 5 𝑝 5 𝑞 5 10 6 𝑝 6 𝑞 4 10 7 𝑝 7 𝑞 3 10 8 𝑝 8 𝑞 2 10 9 𝑝 9 𝑞 1 𝑝 10 ? ? ? ? ? ? ? ? ? ? ? Notice that each of the terms form the Binomial Expansion of 𝑝+𝑞 10 (as per C2). Since 𝑝+𝑞=1, 𝑝+𝑞 10 =1, which shows that the probabilities add to 1 (as we’d hope!) For this reason we call this type of distribution a Binomial Distribution.

FICT Binomial Distribution ! e.g. We threw a coin 𝑛 times. We have a Binomial Distribution when we have: a FIXED number 𝑛 trials. The trials are INDEPENDENT. A CONSTANT probability 𝑝 of success (in each trial) TWO outcomes in each trial, “success” and “failure”. If these conditions are met, a random variable 𝑋 whose outcome is the number of successes is Binomially distributed: 𝑋~𝐵 𝑛,𝑝 𝑃 𝑋=𝑥 = 𝑛 𝑥 𝑝 𝑥 1−𝑝 𝑛−𝑥 So that we’re allowed to multiply the probabilities together from each trial We’re counting heads, so throwing a heads is the “success”. 𝑥 is number of successes. If 𝑛 trials and 𝑥 successes, must be 𝑛−𝑥 failures. FICT

Quickfire Questions 𝑋~𝐵 10,0.3 𝑷 𝑿=𝟒 = 𝟏𝟎 𝟒 𝟎.𝟑 𝟒 𝟎.𝟕 𝟔 ? Show the calculation required to find the indicated probability given the distribution. 𝑋~𝐵 10,0.3 𝑷 𝑿=𝟒 = 𝟏𝟎 𝟒 𝟎.𝟑 𝟒 𝟎.𝟕 𝟔 𝑋~𝐵 10,0.2 𝑷 𝑿=𝟑 = 𝟏𝟎 𝟑 𝟎.𝟐 𝟑 𝟎.𝟖 𝟕 𝑋~𝐵 5,0.1 𝑷 𝑿=𝟐 = 𝟓 𝟐 𝟎.𝟏 𝟐 𝟎.𝟗 𝟑 𝑋~𝐵 20,0.45 𝑷 𝑿=𝟐𝟎 = 𝟎.𝟒𝟓 𝟐𝟎 𝑋~𝐵 20,0.45 𝑷 𝑿=𝟎 = 𝟎.𝟓𝟓 𝟐𝟎 ? ? ? ? ?

Is it Binomially Distributed? Is a Binomial Distribution appropriate as a model? The number of red balls selected when 3 balls are drawn from bag of 15 white and 5 red balls. Some number out of 8 people being left-handed Number of throws on die until 6 obtained Number of girls in family of 4 children ? ? ? ? We have a fixed number 𝑛 trials. Each trial has two possibilities, “success” and “failure”. The trials are independent. There is a fixed probability 𝑝 of success in each trial. No, not fixed. This is known as a ‘Geometric Distribution’ (which we won’t cover)         Technically the probability of having a girl increases if you previously had a girl, and vice versa. But the probability is still close to 0.5, so Binomial Distribution is appropriate. Usually. But in my story, genetics has an influence on handedness. Only if balls drawn with replacement.  Only if balls drawn with replacement, 𝑝= 1 4   

Test Your Understanding 𝑋~𝐵 12, 1 6 What is 𝑃(𝑋=2)? 𝑷 𝑿=𝟐 = 𝟏𝟐 𝟐 𝟏 𝟔 𝟐 𝟓 𝟔 𝟏𝟎 =𝟎.𝟐𝟗𝟔 What is 𝑃(𝑋≤1)? 𝑷 𝑿≤𝟏 =𝑷 𝑿=𝟎 +𝑷 𝑿=𝟏 = 𝟓 𝟔 𝟏𝟐 + 𝟏𝟐 𝟏 𝟏 𝟔 𝟏 𝟓 𝟔 𝟏𝟏 =𝟎.𝟑𝟖𝟏 I have a bag of 2 red and 8 white balls. 𝑋 represents the number of red balls I chose after 5 selections (with replacement). How is 𝑋 distributed? 𝑿~𝑩 𝟓,𝟎.𝟐 Determine the probability that I chose 3 red balls. 𝑷 𝑿=𝟑 = 𝟓 𝟑 𝟎.𝟐 𝟑 𝟎.𝟖 𝟐 =𝟎.𝟎𝟓𝟏𝟐 Q1 ? ? Q2 a ? b ? (If you get these quickly, go on to Exercise 1B)