Show
Introduction Probability is the likelihood or chance of an event occurring.
For example, the probability of flipping a coin and it being heads is ½, because there is 1 way of getting a head and the total number of possible outcomes is 2 (a head or tail). We write P(heads) = ½ .
This video is a guide to probability. Expressing probability as fractions and percentages based on the ratio of the number ways an outcome can happen and the total number of outcomes is explained. Experimental probability and the importance of basing this on a large trial is also covered. Single Events Example There are 6 beads in a bag, 3 are red, 2 are yellow and 1 is blue. What is the probability of picking a yellow? The probability is the number of yellows in the bag divided by the total number of balls, i.e. 2/6 = 1/3. Example There is a bag full of coloured balls, red, blue, green and orange. Balls are picked out and replaced. John did this 1000 times and obtained the following results:
a) What is the probability of picking a green ball? For every 1000 balls picked out, 450 are green. Therefore P(green) = 450/1000 = 0.45 b) If there are 100 balls in the bag, how many of them are likely to be green? The experiment suggests that 450 out of 1000 balls are green. Therefore, out of 100 balls, 45 are green (using ratios). Multiple Events Independent and Dependent Events Suppose now we consider the probability of 2 events happening. For example, we might throw 2 dice and consider the probability that both are 6's. We call two events independent if the outcome of one of the events doesn't affect the outcome of another. For example, if we throw two dice, the probability of getting a 6 on the second die is the same, no matter what we get with the first one- it's still 1/6. On the other hand, suppose we have a bag containing 2 red and 2 blue balls. If we pick 2 balls out of the bag, the probability that the second is blue depends upon what the colour of the first ball picked was. If the first ball was blue, there will be 1 blue and 2 red balls in the bag when we pick the second ball. So the probability of getting a blue is 1/3. However, if the first ball was red, there will be 1 red and 2 blue balls left so the probability the second ball is blue is 2/3. When the probability of one event depends on another, the events are dependent. Possibility Spaces When working out what the probability of two things happening is, a probability/ possibility space can be drawn. For example, if you throw two dice, what is the probability that you will get: a) 8, b) 9, c) either 8 or 9?  a) The black blobs indicate the ways of getting 8 (a 2 and a 6, a 3 and a 5, ...). There are 5 different ways. The probability space shows us that when throwing 2 dice, there are 36 different possibilities (36 squares). With 5 of these possibilities, you will get 8. Therefore P(8) = 5/36 . b) The red blobs indicate the ways of getting 9. There are four ways, therefore P(9) = 4/36 = 1/9. c) You will get an 8 or 9 in any of the 'blobbed' squares. There are 9 altogether, so P(8 or 9) = 9/36 = 1/4 . Probability Trees Another way of representing 2 or more events is on a probability tree. Example There are 3 balls in a bag: red, yellow and blue. One ball is picked out, and not replaced, and then another ball is picked out. The first ball can be red, yellow or blue. The probability is 1/3 for each of these. If a red ball is picked out, there will be two balls left, a yellow and blue. The probability the second ball will be yellow is 1/2 and the probability the second ball will be blue is 1/2. The same logic can be applied to the cases of when a yellow or blue ball is picked out first. In this example, the question states that the ball is not replaced. If it was, the probability of picking a red ball (etc.) the second time will be the same as the first (i.e. 1/3). This video shows examples of using probability trees to work out the overall probability of a series of events are shown. Both independent and conditional probability are covered.
The AND and OR rules (HIGHER TIER) In the above example, the probability of picking a red first is 1/3 and a yellow second is 1/2. The probability that a red AND then a yellow will be picked is 1/3 × 1/2 = 1/6 (this is shown at the end of the branch). The rule is:
The probability of picking a red OR yellow first is 1/3 + 1/3 = 2/3. The rule is:
On a probability tree, when moving from left to right we multiply and when moving down we add. Example What is the probability of getting a yellow and a red in any order? This is the same as: what is the probability of getting a yellow AND a red OR a red AND a yellow. P(yellow and red) = 1/3 × 1/2 = 1/6 P(red and yellow) = 1/3 × 1/2 = 1/6 P(yellow and red or red and yellow) = 1/6 + 1/6 = 1/3 21 Jan 2018 In today's post, I want to (re-)introduce some simple math that most people learn in high school, but forget quicly: permutations and combinations. Permutations and combinations allow you to answer questions like: In how many ways can I order a set of distinct items? If I take K of those items, how many combinations of them are there? Then we will go a bit further and see how to deal with repetition (putting back an item after selecting it), multisets (sets in which there are multiple copies of the same item, like a pit that would hold 4 red balls, 3 blue balls and 2 green balls) and picking from different sets. Sets of Distinct ItemsImagine you have a set of distinct items. This could be a deck of cards, or a pit containing uniquely numbered balls. We will look at the number of ways to pick items out of this set, with and without taking ordering into account. One basic rule: Once an item has been picked from the set, it can't be picked again. We will look at three questions:
Let's now give the formula and intuition for how to answer these questions. PermutationsThere are n! ways to order the elements of a set of size n. n! means the factorial of n. The factorial of 3 is 3 * 2 * 1 = 6. The factorial of 5 is 5 * 4 * 3 * 2 * 1 = 120, etc. The factorial of 0 is 1. Intuition: Imagine we have to order 3 balls {1, 2, 3}. First we pick one ball and we have three to choose from, so that's three possibilities. Next we pick another ball, and there are only two left to choose from. This means that for each choice we made when picking the first ball, there are two further possibilities for the second ball, so that's 3 * 2 = 6 possibilities. Finally there is only one ball left to pick: we have no choice. This means there are 3! = 3 * 2 * 1 = 6 ways to order the three balls. k-PermutationsThe are P(n, k) = n! / (n-k)! ways to order k items picked from a set of size n. For other mathematical notations, see here. Intuition: We start by considering all n! permutations of the set. But we are only interested in the k first items of these permutations. Amongst all permutations, there are multiple that start with the same k items. How many of these are there? Well, the permutations that have the same first k items must have a different permutations of their n-k last items. With the formula for permutations, we know there are (n-k)! ways to form permutations of a set of n-k items. So by dividing the total number of permutations (n!) by the number of permutations of the last n-k items ((n-k)!), we only keep one permutation per possible k-permutation, and that is our answer. (k-)CombinationsThe are C(n, k) = n! / (k! * (n-k)!) ways to pick k items from a set of size n. For other mathematical notations, see here. Intuition: We start from the n! / (n-k)! k-permutations of the set. These have the right size, but they take ordering into account. This means the combinations are counted multiple times, with different ordering. How many ordering of each combination are there? A combination is just a subset of size k of the whole set. With the formula for permutations, we know there are k! ways to order the items of this set. So by dividing the number of k-permutations (n! / (n-k)!) by the number of permutations (k!) for each combination, we end up getting the number of combinations! Sets of Distinct Items, with RepetitionBy repetition, we mean that we change the basic rule of the last section: once a item has been picked from the set, we put it back and are free to pick it again, as many times as we like. This time, looking at the number of permutations of a whole set doesn't make sense anymore, as we put back each item after picking it. Let's look at the other two questions in the presence of repetitions: k-Permutations with repetitions (or n-tuples)There are n^k ways to order k items from a set of size n, with repetition. Intuition: each time we pick an item, we have n possibilities, and these possibilities just multiply for each item we pick. For instance, for a set of two balls {1, 2}, there are 2^3 = 8 ways to pick 3 balls: 111, 112, 121, 122, 211, 212, 221, 222. There are CC(n, k) = C(n+k-1, k) = (n+k-1)! / (k! * (n-1)!) ways to pick k items from a set of size n, with repetition. For the proper mathematical notation, see here. Intuition: This formula is rather harder than the rest, and so the intuition is a bit more indirect. The idea is that each k-multicombination can be seen as a sequence of k stars and n-1 separating bars. Basically, the bars define n separated intervals (for instance 1|2|3). Each interval corresponds to an item in the set, and the number of stars (zero or more) in that interval corresponds to the number of times that item was picked from the set. So for instance, ***|*||* represents a way to pick 5 (k) items from a set of 4 (n) items. In this particular solution, the first item is picked thrice, the third not at all, and the other two only once. So if we want to know the number of k-multicombinations, the problem reduces to that of counting the number of such sequences. The sequences will have length n+k-1, and there are as many sequences as ways to pick the position of the k stars within the sequence. That is a combination! In particular, it is C(n+k-1, k), and the rest is just applying the combination formula. Picking From Multiple SetsBefore we get to the somewhat harder problem of picking in sets that contains multiple copies of the same items, let's see what happens when we pick from multiple distinct sets. We want to answer questions like: how many ways are there to pick 3 balls from my first pit containing 10 balls, and 4 balls from my second pit containing 8 balls. It's very simple: since picking from each set is independent, we just compute the two combinations, then multiply them: we can pair any combination from the first set with a any combination from the second set. The same works for permutations, and we can even mix the two kinds. Just remember: when two number of possibilities are independant (all combinations of the two set of possibilities are valid), you can just multiply them. MultisetsIn what follows, we consider multisets: sets that have repeated items. We always consider a multiset of size n containing m kind of items, such that there are k1, k2, ..., km copies of the first, second, ..., last item. Note that k1 + k2 + ... + km = n. Permutations of MultisetsThere are n! / (k1! * k2! * ... * km!) ways to order the elements of a multiset. This formula is also called the "multinomial coefficient". Intuition: There would be n! ways to order the items of the multiset, if we considered each item to be unique. Let's consider the first kind of items (let's say "red balls"). All configurations that have red balls in the same positions are identical. For each of our n! permutations, how many have the red balls in the same position, except with different red balls? This is equivalent to asking to asking how many ways there are to distribute the red balls in the "red slots". This is the same as asking how many ways there are to order all the red balls, which is a simple permutation: k1! ways. Now we know there are n!/k1! permutations in which we do not distinguish between the red balls, but still distinguish between all the other balls individually. To obtain, the final result, the same process must be performed for each kind of ball, successively divising until we obtain the final formula: n! / (k1! * k2! * ... * km!). k-Permutations of MultisetsThis is much more complicated than what we'd seen so far. A formula can be derived, but it is unwieldy. I will still try to give some intuition, but if you're after easy to use formulas, feel free to skip this. Observe that if we pick k items from our multiset such that k < k1, k < k2, ..., k < km, then the problem is essentially that of k-permutations with repetition, and there are thus m^k such permutations. The reason is that we can never run out of any kind of item. When this condition doesn't hold anymore, we would like to still start from m^k hypothetical permutations, but remove from this total the permutations where we picked more of a kind of item than there were in the multiset. This seems easy, but there is a problem: some of the permutations that have too many of the first kind of items may also have too many of the second kind of item. We should not remove them twice! Let's call TM1, TM2, ..., TMm the number of permutations where there are Too Many of the 1st, 2nd, ..., mth item. Computing these values is not trivial: for each TMi, we have to sum all k-combinations of x items amongst k, for all x such that ki < x <= k. When we have two kind of items, the simplified formula is: m^k - TM1 - TM2 + TM(1,2). The meaning of TM(1,2) here is "number of of permutations where there are too many of the first kind of item AND too many of the second times of item". Those permutations are actually substracted twice (once via TM1, once via TM2) and so need to be added back. What about three kind of items? It's m^k - TM1 - TM2 - TM3 + TM(1, 2) + TM(1, 3) + TM (2, 3) - TM(1, 2, 3) We now have three combinations of item kinds to consider. But after adding back these permutations, we still aren't done! Consider the permutations where we have too many of all three kind of items. We have removed them thrice (TM1, TM2, TM3) and added them back thrice (TM(1, 2), TM(1, 3), TM(2, 3)), so we need to remove them one final time (TM(1, 2, 3)). The more kinds of items you have, the crazier it gets. Not to mention that all these TM values are hairy to compute by themselves. Still easy stuff for a computer to do, but don't try this by hand. This principle of removing and adding back duplicates is known as The Inclusion-Exclusion Principle k-Permutations of Multisets with RepetitionThis is actually easy! Because repetition is allowed, it doesn't matter how many of each kind of item there is. The answer is thus the same as the number of k-permutations with reptition for a set that would have only one of each kind of item: k^m (where m is the number of kinds of items). However, there is a pitfall: while all k^m k-permutations are possible, they aren't all equiprobable — they don't all have the same probability to be selected randomly. The probabilities for each k-permutation will depend on the number of items of each kind. k-Combinations of MultisetsThis has just the same problem as counting the number of k-permutations of a multiset, the only difference being we do not consider ordering. The reasoning is much the same, so I shall say no more of it. k-Combinations of Multisets with RepetitionThis, in turn, can be obtained in much the same way as the number of k-permutations of multisets with repetition. Namely, there are CC(m, k) = C(m+k-1, k) = (m+k-1)! / (k! * (m-1)!) such k-combinations. And again, not all of them are equiprobable. ProbabilitiesCombinations and permutations are often used for deriving the probability that a class of event will occur. The general recipe goes something like this:
So to take our example, there are C(52, 3) ways to draw three cards out of a full 52-cards deck. There are C(13, 3) ways to pick three hearts cards from all heart cards in the deck, so that's also the number of scenarios in which we pick three cards from our full deck. So there is a C(13, 3) / C(52, 3) = 0.013 = 1.3% probabability to pick 3 heart cards from a full deck. Remember you can multiply scenarios if they are independent in order to get the total number of scenarios. This corresponds to an AND within a scenario. For instance, the probability to draw three red cards and three black cards from a full deck. (Answer: C(26, 3)^2 / C(52, 6)) You can also add scenarios when they are independent. This corresponds to an OR within the scenario. Example: There are 12 balls in a bag, 3 of them are red, 4 of them are green and 5 of them are blue. We randomly take out 3 balls from the bag at the same time. What is the probability that all three balls are of the same colour? (Answer: (C(5, 3) + C(4, 3) + C(3, 3)) / C(12, 3). Beware of scenarios that are not independant. For instance, if you search the probability to draw four cards, three of which are red, and one of which is a figure, you can't just multiply the probability to pick three red cards and the probability to pick a figure: the figure might be one of your red cards! There are ways around these issues, and we already mentionned the The Inclusion-Exclusion Principle. I'm afraid that in intricate cases, there is no substitute for some thinking. More information:
|
In how many ways can we pick any number of balls from a pack of 3 different balls
Postagens relacionadas
direito autoral © 2024 estudarpara Inc.
