Probability Question

Question

The sample space S of a random experiment is {1, 2, 3, 4, 5, 6, 7, 8}. Assuming that the outcomes of this experiment are equally likely, answer the question given below. 

X is another event with P(X) = 0.7. Write two events that X could be

Answer 1

There are 70 combinations of quartets of these 8 numbers, or, in other words, 70 different ways of grouping 4 outcomes from S. They all have the same probability of being selected as a group because each member of S has the same probability of selection. These are combinations of the possible outcomes of simply picking 4 numbers from 1 to 8 so that once drawn, a number cannot be returned to the pool. After a group of 4 has been drawn, then all the numbers are returned to the pool for the next group selection in this experiment.

If we add together the numbers in each quartet we get a particular sum, associating each quartet with a sum. For example, 1+2+3+4=10 and 5+6+7+8=26. So the sum is in the range 10 to 26, and the sums are not usually unique. For example: 1+2+3+8=1+3+4+6=2+3+4+5=14. So, given a sum we can identify how many quartets have that sum. Or, we can find out how many have less or more than this sum. That gives us a probability based on the number of occurrences divided by the total number of possibilities.

If we can find a sum that is shared by 49 quartets then the probability of selecting that sum will be 49/70=7/10=0.7. Alternatively, we can count the number of quartets that are below or above that sum. This will be one of the requirements of the experiment.

But it doesn't have to be a sum. We can use other operators than plus. In fact, between 4 numbers we can have three operators, for example, 1*2+3*4=14, or 1*2*3*4=24. So two possible events for X will simply be the result of applying different operators. If we take the sample space for the operators as { + - * } then there will be 6 permutations of these, if all operators are to be applied in order, to ordered elements of each quartet. The elements can be ordered in ascending or descending order and the results will usually be different. The normal priority rules would apply (multiplication takes priority over addition and subtraction). All that's needed is a particular result (< X, ≤ X, = X, ≥ X, > X), so that the number of occurrences is 49 out of the possible 70 outcomes. Then P(X)=49/70=0.7.

The question doesn't ask for specific X, just possibilities for X.

If I can, I'll identify specific events for X...when I have time!

 

Comments

Thank you Mr. Rod. But how can you say There are 70 combinations of quartets of these 8 numbers. And why did you get 4 numbers for a group?

 

This is the full question

The sample space S of a random experiment is {1, 2, 3, 4, 5, 6, 7, 8}. Assuming that the outcomes of this experiment are equally likely, answer the questions given below.

(i) A is a simple event of the above experiment. Write down all the events that A could be.

   Answer : {1}, {2}, {3}, {4}, {5}, {6}, {7}, {8}

(ii) For each of the above events, find P(A).

     Answer : For each event P(A)=1/8

(iii) B is a compound event of the above experiment consisting of 4 elements. Write one event that B could be.

     Answer : {1,2,3,4}

(iv) Find P(B) and P(B').

      Answer : P(B)= 4/8=1/2             P(B')= 1 - P(B)= 1/2

(v) X is another event with P(X) = 0.7. Write two events that X could be.

     Answer , This part is the problem for me

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I couldn't access the system so I've only just read your complete question. Nevertheless I had prepared an answer below.

OK. There are 70 combinations of selecting 4 from 8 different objects because the formula for combinations is 8*7*6*5/(1*2*3*4)=70. The set of outcomes { 1 2 3 4 5 6 7 8 } in your question, is the set of objects referred to. By using the selection of a quartet of 4 different outcomes to form a 4-digit number (with the digits in ascending order), we can write down all possible 4-digit numbers: 1234, 1235, 1236, 1237, 1238, 1345, 1346, 1347, etc. 1234 represents drawing 4 numbers from the sample space set of 1 to 8. These numbers are unique and there are 70 of them. There are 35 beginning with 1, 20 beginning with 2, and so on. In fact from 1234 to 2467 there are 49 numbers in this set of 70. Therefore, as an example, we can use the probability P(x ≤ 2467) where x is an element of this set of 70. Since we know there are 49 numbers no bigger than 2467 the probability of selecting at random a 4-digit number (made up of different digits taken from 1 to 8) is 49/70=7/10=0.7.

 

So we've found a P(X)=0.7 exactly by stipulating that X is defined as x ≤ 2467.

That's one set. To find a different X we start at the highest number 5678 and count backwards till we have included 49 numbers. That takes us to 1358, so X is now x ≥ 1358 and P(x ≥ 1358)=0.7. So now we have two values of X (two conditions or event requirements) that give us the required probability.

 

In the answer you will have seen so far, I suggested perhaps adding the digits of the quartet would work. But when I investigated, it's only possible to approximate to 0.7 (I got 46/70=0.66 approximately for X=sum of digits ≤ 19. P(sum≤19)=0.66.

 

I don't know whether this is what the question is asking for, but I offer my solution as above as a possible answer.

 

So the experiment referred to consists first of drawing 4 numbers out of 8. The numbers are then arranged in numerical order to form a 4-digit number. This gives us a sample space for a different event. There are 70 elements in this set, 49 of which are less than or equal to 2467 and 49 are greater than or equal to 1358.

 

Why did I pick 4 numbers and not 3 or something else? Because somehow I needed to create a probability of 7/10 and then I discovered that 70 happened to be the number of combinations of 4 out of 8. Then I realised that 49/70=7/10 so all that was needed now was to invent a situation that would give me a result of 49/70. That's how I did it!

 

I hope this explanation is clear.

 

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Thank you so much Mr.Rod.

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This is a list of the 70 numbers with the limits in the text highlighted in bold print.