# Puzzling Permutations& Combinations and Probability questions

• chswallow
In summary: Similarly, for 2(ii), consider each card in turn. Doesn't matter which the first one is, right? Now, what's the chance the second will be different? Next, assuming the first two are different, what's the chance that the third will be different from the first two, and so on. Since you need all these things to happen to end up with all different ones, and since they're independent, multiply those chances together.So, the probabilities for 3(i) are:3(i) = (1/6)*(1/2)*(1/3) = 1/363(i) = (1/6)*(1/2
chswallow
Hi all, I have a set of questions, with answers in brackets at the end of each of them. I really don't know how to solve them, and have spent a whole morning trying to figure them out. Please help me! (answer one/all of them!)

1. In a game of chance, each player throws two unbiased dice and his score is the difference between the numbers on the dice. Anthony and Bob play against each other and the player whose score is at least four more than his opponent's score wins. FInd the probability that neither player wins. (74/81)

2. Every person belongs to one of the four blood groups: O, A, B, AB. Proportions of the population in a city belonging to these blood groups are 0.46, .40, .11, .03 respectively. If 3 people are chosen at random, what is the probability that they all belong to different blood groups? (0.172)

3. A fast food restaurant gives away a free action figure for every child's meal bought. THere are 5 action figures and each figure is likely to be given awe with a child's meal. A customer intends to collect all 5 different action figures by buying child's meals.
i) Find the probability that the first 4 child's meals bought by the customer all had different actin figures. (0.224)
ii) At a certain stage, the customer has collected 4 of the 5 action figures. Find the least number of child's meals needed so that the probability of the customer completing the set is larger than 0.95. (14)

4. The probability that a soccer team wins any match is 0.5, and the probability that it loses any match is 1/6. Three points are awarded for a win, one point for a draw and no point for a defeat. If the team plays five matches, find the probability that the team:
i) wins exactly 3 matches given that it wins the first match (3/8)
ii) wins exactly one match given that it obtains exactly five points (45/49)

5. Eleven cards, bearing the letters of the word INDEPENDENT, are placed in a box. 3 cards are drawn at random without replacement. Calculate the probability that
i) exactly two of the cards bear the same letters (16/55)
ii) all 3 cards bear different letters (116/165)
iii) the 3 cards bear the letters D, E, N in that order. (2/110)
iv) the three cards bear the letters D, E, N in any order. (6/55)
v) if the first two cards drawn are E and N respectively, and the drawing of the cards will continue without replacement until a card bearing D is obtained, find the least value of n such that P(at most n more cards are drawn to get a D)> 0.8 (5)

THANK YOU!

Hey chswallow and welcome to the forums.

Lets look at the first question.

In the first question you need to find the probability that P(|Y1 - Y2| < |X1 - X2| + 4) OR P(|X1 - X2| < |Y1 - Y2| + 4) where X1,X2 are the values of the dice rolls for player 1 and Y1,Y2 are the values for the dice rolls for player 2 (both are random variables).

Now you know that X1,X2,Y1,Y2 have a distribution of discrete uniform from 1-6 with probability P(X = x) = 1/6 for x = 1,2,3,4,5,6.

Can you use this to expand out the probability functions firstly for |X1 - X2| and |Y1 - Y2| and later to find out P(|X1-X2| < |Y1-Y2|+4) (and also for the other way around)?

For several of these, I don't think there's anything clever to be done. You just have to list the ways the event can happen (or ways it doesn't happen, if that looks like being less) and calculate the probability of each.
For 3(i), consider each action figure in turn. Doesn't matter which the first one is, right? Now, what's the chance the second will be different? Next, assuming the first two are different, what's the chance that the third will be different from the first two, and so on. Since you need all these things to happen to end up with all different ones, and since they're independent, multiply those chances together.

## 1. What is the difference between permutations and combinations?

Permutations and combinations are both methods of counting the number of ways that a group of objects can be arranged or selected. The main difference is that permutations take into account the order of the objects, while combinations do not. In other words, permutations are concerned with arranging objects in a specific order, while combinations are concerned with selecting objects regardless of their order.

## 2. How do I calculate the number of possible permutations or combinations?

The formula for calculating permutations is n! / (n-r)!, where n is the total number of objects and r is the number of objects being selected. For combinations, the formula is n! / (r!(n-r)!). It is important to note that the exclamation mark (!) represents the factorial function, which means multiplying a number by all the numbers before it. For example, 5! = 5 x 4 x 3 x 2 x 1 = 120.

## 3. When should I use permutations and when should I use combinations?

Permutations should be used when the order of the objects matters, such as when arranging a sequence of events or creating a password. Combinations should be used when the order does not matter, such as selecting a group of people for a committee or choosing toppings for a pizza.

## 4. How do probability and permutations/combinations relate to each other?

Probability is the likelihood of a certain event occurring, while permutations and combinations are used to calculate the number of possible outcomes. In some cases, probability can be calculated by dividing the number of desired outcomes by the total number of possible outcomes, which can be determined using permutations or combinations.

## 5. Are there any common mistakes to watch out for when working with permutations and combinations?

One common mistake is forgetting to use the factorial function in the formulas. Another mistake is confusing permutations and combinations and using the wrong formula for the given scenario. It is important to carefully read the question and determine whether the order of objects is important or not before applying the appropriate formula.

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