Psychology 210 Statistical Methods February 1, 2006 Probability calculations with M&Ms Get a packet of M&Ms from me. Open it up and lay the candies out on a piece of notebook paper. Don’t eat them yet. Count up the number of candies in your bag (or to use the language of statistics, determine the n of your sample). Please don’t blame me if your neighbor has a larger sample than you. Life is sometimes unfair. n = ____________ Now count out the frequencies of each color of M&M and fill in the frequency column of the table below. Color Frequency Percent Probability Red Orange Yellow Green Blue Brown Total Calculate the percent of M&M’s of each color (express as a percentage; e.g., 20%). These should all add up to 100. The probability of an event, A is the likelihood that it will occur, relative to all of the other events that can occur. nfEP =)( where: f is the frequency of event class E n is the total frequencies of all events in the distribution Note: This is an example of empirical probability - probability based directly on observations.If we consider each M to be an event that we can draw at random from a sample, we can calculate the probabilities of drawing a candy of each color using the above formula. For example, the probability of a red M&M should be equal to: )()(Re)(ReanyndfdP = Fill in the Probability column using fractions (e.g., 3/18) As you do, notice the relationship of this column to the Percent column. If this doesn’t make sense, stop and consider the denominator in the above formula for probability. The addition rule In order to calculate the probability that one of two mutually exclusive events will happen, one should add the probabilities of each of those happening independently. P(A or B) = P(A) + P(B) Red and Blue are both “new” colors to some of the older people in the room. That is, they were introduced (or in the case of red, reintroduced) later than the others. Calculate the probability of a new color using the following formula. P(New) = P(R or Bl) = P(R) + P(Bl) = ________________________________________ Compare this to the percentage of all M&M’s that are either red or blue from your table. They should be the same, no? Now try these other problems: The probability of a “warm” color (red, orange, or yellow) P(Warm) = ______________________________________________________________ The probability of a “cool” color (blue or green) P(Cool) = _______________________________________________________________ The probability of a primary color (Yellow, blue or red) P(Primary) = ____________________________________________________________ The probability a color that starts with the letter B P(LetterB) = _____________________________________________________________ The probability of blue, green, orange, yellow, brown or red P(Bl or G or O or Y or B or R) = ____________________________________________Sampling without replacement If we make multiple draws from a limited sample (rather than an infinite vat of M&Ms) then I should no longer use the same, theoretical probabilities for each draw. This is because once one has drawn and eaten an M&M, the number remaining in the sample has changed and consequently, so have actual probabilities. For example, if 4 out of 20 candies were red, and I draw one (with probability 4/20 or .20) the probability of subsequently drawing another red has changed. There are now 3 reds left out of 19 total, for a probability of 3/19, or .16 For your sample, calculate the probability of drawing the following. Remember your calculation for P(B) is after A has been drawn. The table may help. Situation P(A) P(B) P(A,B) P(G,R) P(R,R) P(O,O) P(Y,Y) P(G,G) P(Bl,Bl) P(Br,Br) Now, figure the probability of any “double”. In other words, the probability that your first two random draws will be the same color (as in the last 6 rows of your table) Finally, calculate the probability of randomly drawing a blue, then a red, then a brown. Again, this should be for your specific sample.Conditional Probability What you’ve just been doing involved calculating conditional probabilities. Specifically, the probability for the second draw was dependent (or conditional) on the first. This example should clarify the idea: The probability of drawing a red on the second draw would be different, depending on whether your first draw was a blue or a red candy. Specifically, it would be smaller in the latter case. So the first draw places a condition on the second. Another way to phrase this is, “what is the probability of drawing a red, given that I just drew a red? There are several other ways to use conditional probabilities. They can be used to make more accurate guesses about probabilities if given further information. If you knew that I always chose cool colors (blue or green), you could estimate the likelihood of my selection more accurately than if you didn’t have that knowledge. In a conditional probability, the probability of A given B (or AB) is equal to: )()()|(APAandBPABP= This only looks at those possibilities that are included in the given condition. Use this formula to calculate the probability of green given a cool color: P(Gr | Cool) = ___________________________________________________________ Another friend only likes colored Ms and eschews brown ones. Find the probability of a blue given not brown, or: =)|( BrBlP _____________________________________________________________ How about the probability of red given that it’s a “new color”: P(R | New) = ____________________________________________________________ * Note: the above formula is the general formula for conditional probability. If it is the case that all B are A (as is the case for these problems), it simplifies to the more logical: )()()|(APBPABP =Classical Probability and The Multiplication rule M&M Mars has invested a great deal of research into finding the most appetizing distribution of colors to put in bags of M&M candies. The following are the target percentages for each color that the manufacturer tries for. 10% Blue 10% Green 10% Orange 20% Yellow 30% Brown 20% Red First, compare these to the percentages of your sample. They’re probably not spot-on because of random variation, but most likely they are very close. If you were reaching into a bag of M&Ms without knowing the contents these would be your best guesses