Guide :

# Count the number of elements in the event

A state lottery game consists of choosing one card from each of the four suits in a standard deck of playing cards. (There are 13 cards in each suit) Count the number of elements in the event than an ace, a king, a queen, and jack are chosen.

## Research, Knowledge and Information :

### Counting - Department of Statistics

... it is easier to find the number of elements in a set by counting each ... that can be helpful in counting: Count the elements in a set larger ... event; factorial ...

### count of events per datetime R - Stack Overflow

count of events per datetime R. ... and then you can use a number of aggregation functions to count the events per day: ... The sum of the elements of an array

### shell script - Count number of elements in bash array, where ...

... Is there built-in bash method to count number of elements in ... count=\${#!EXPR1} # ERROR: bash: !EXPR}: event not found 3. count ... Count number of elements ...

### Count Property | Microsoft Docs

Gets the number of entries in the event log (that is, the number of elements in the ... to use the Count property to ... of the Event Log ...

### c++ - count specific number of elements - Stack Overflow

count specific number of elements. ... Count number of elements in an array using Javascript. 3. ... saved' event Earliest ...

### Answers for Number Of Elements Int The Event Spac... - OpenStudy

Answers for Number Of Elements Int The Event Space Of Rolling A Sum Of 5 With Two Dice ... Answers for Number Of Elements Int The Event Space Of Rolling A Sum Of 5 ...

### A state lottery game consists of choosing one card from each ...

Count the number of elements in the event that an ace, a king, a queen, and a jack are chosen. ... So the number of elements in the event is 4 * 3 * 2 * 1 = 24.

### PHP count() Function - W3Schools

PHP count() Function PHP Array ... function returns the number of elements in an array. Syntax. count ... Does not count all elements of multidimensional ...

## Suggested Questions And Answer :

### Probability Question

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!

### Count the number of elements in the event

Let's call the valued cards royal. In a pack of 52 cards there will be 16 royals. The odds of picking one of them is 16 in 52. That leaves 51 cards containing 15 royals. So the odds of picking another royal is 15 in 51. So we continue until only 37 remain, one of which is the remaing royal. The odds of picking all 16 royals in any order is therefore 16/52*15/51*...1/37 to 1 against. The number of events is the reciprocal of this which is 52C16 in nCr format. This evaluates to 10,363,194,502,115. Go for it - you can't lose!! Not!

### calculate number of subsets for the set [5,17,3,19,6]

I'll approach this answer assuming you have a calculator with an nCr button (we have 5 members in the set; if we want to know how many unique sets with 2 numbers per set then we enter for nCr 5C2 (5 choose 2). 5C2= 10. This tells us there are 10 combinations of 2 in a set of 5.  In terms of position, the positions of the numbers are position 1,2,3,4,5.  Combinations of 2 include 1-2, 1-3, 1-4, 1-5, 2-3, 2-4, 2-5, 3-4, 3-5, 4-5; or 10 pairs. To find the total number of subsets you need to find the number of subsets with 1 element, the number of subsets with 2 elements, ... 3 elements, 4 elements and 5 elements. The total nuber of subsets of a 5 element set is 5C1 + 5C2 +5C3 + 5C4 +5C5  ; you normally add one more set to the answer to represent the empty set. 5C1 =5 5C2 = 10 5C3 =10 5C4 = 5 5C5 =1 The total number of subsets is 32, or 33 if you count the empty set

### What is discrete and continuous variable?

A discrete variable is one that is restricted (usually) to whole numbers or integers, while a continuous variable can take any real value. Examples of discrete variables  are frequencies of events, where the event can only happen a whole number of times. These variables often appear in statistics when associating a probability to a number of events occurring. They are also restricted to positive whole numbers. Another example of discrete variables would be a problem in making up a sum of money using coins and notes. The solution would only contain discrete numbers of the various denominations of currency. They would be whole, positive integers. An example of a continuous variable would be a graph of a relationship or an equation containing an independent variable and a dependent variable. Their relationship may be represented by a continuous curve.  Another example of a continuous variable would be temperature which varies continuously, even though it may be restricted to a range. Time is also regarded as continuous. Anything which is countable is discrete. If it can't be counted, it's probably continuous.

### shortcut to finding the median without a computer

Question: shortcut to finding the median without a computer. The median is the middle value. If you have an array of values, then the median is that value that separates the upper half of the array from the lower half. If the number of elements in your array A[ ] is odd, i.e. N = 2n+1, then the median, M = A[n+1] If the number of elements in your array A[ ] is even, i.e. N = 2n, then the median, M = (A[n-1]+A[n+1])/2 i.e. M is the mean of the two middle values. If you don't want to use a computer, then you will have to visually inspect the arrray. Once you know the number of elements in the array, N, then you will have to manually count along until you reach the middle element, which is your median (or two middle ements, the mean of which is your median)

### A Question on Probability

The number of Akron machine customers is 160+20=180. The number of Wheeling machine customers is 50+150=200. Total number of customers is 180+200=380. The probability of a random customer being an Akron customer is 180/380=9/19. So the probability of being a Wheeling customer is 1-9/19=10/19 (=200/380). So (a) P(A)=9/19 or 0.4737 or 47.37%. The meaning of Rc hasn't been defined so I take it to be the complement of R, in other words, the set of customers not having the in-home repair warranty. Similarly, Ac represents the Wheeling customers. Out of 380 customers, (20+50)=70 took up the offer of the in-home repair, so that's (b) P(R)=70/380, 0.1842 or 18.42%. The probability P(Rc) of no in-home repair warranty is 1-70/380=310/380=0.8158, 81.58%. (c) The probability of A intersection R, that is, being an Akron customer and having the repair warranty is 20/380=1/19=0.0526, 5.26%. (d) To find the probability of being an Akron customer or having the warranty, we need to count the customers matching these requirements. 180 Akron customers + 50 Wheeling customers having the warranty=230, so the probability is 230/380=23/38=0.6053, 60.53%. (e) The total of Wheeling customers, some of whom have the warranty, and Akron customers having the warranty is 200+20=220. This gives the probability 220/380=11/19=0.5789, 57.89%. (f) 180 Akron customer Service + 150 Wheeling customers with no warranty = 330, making the selection probability 330/380=0.8684, 86.84%.

### what is the probability of winning 6 numbers out of 40 ?

Picking 6 numbers out of 40, where you make exactly 6 picks to get those 6 winning numbers, where all 40 possible numbers are different, picked numbers can't be repeated, and the order in which the numbers are picked doesn't matter: P(first picked number is a winner) = 6/40 P(2nd picked number is a winner) = 5/39 P(3rd picked number is a winner) = 4/38 P(4th picked number is a winner) = 3/37 P(5th picked number is a winner) = 2/36 P(6th picked number is a winner) = 1/35 P(all 6 of the above events happen) = 6/40 * 5/39 * 4/38 * 3/37 * 2/36 * 1/35 P(all 6 of the above events happen) = (6 * 5 * 4 * 3 * 2 * 1) / (40 * 39 * 38 * 37 * 36 * 35) P(all 6 of the above events happen) = 34! * 6! / 40! P(all 6 of the above events happen) = "40 choose 6" P(all 6 of the above events happen) = 40 nCr 6   (on a calculator) P(all 6 of the above events happen) = 0.000000260526576  (about 1 in 3.8 million)

### 6 winning number out of 49

Picking 6 numbers out of 49, where you make exactly 6 picks to get those 6 winning numbers, where all 49 possible numbers are different, picked numbers can't be repeated, and the order in which the numbers are picked doesn't matter: P(first picked number is a winner) = 6/49 P(2nd picked number is a winner) = 5/48 P(3rd picked number is a winner) = 4/47 P(4th picked number is a winner) = 3/46 P(5th picked number is a winner) = 2/45 P(6th picked number is a winner) = 1/44 P(all 6 of the above events happen) = 6/49 * 5/48 * 4/47 * 3/46 * 2/45 * 1/44 P(all 6 of the above events happen) = (6 * 5 * 4 * 3 * 2 * 1) / (49 * 48 * 47 * 46 * 45 * 44) P(all 6 of the above events happen) = 43! * 6! / 49! P(all 6 of the above events happen) = "49 choose 6" P(all 6 of the above events happen) = 49 nCr 6   (on a calculator) P(all 6 of the above events happen) = 0.0000000715112384   (about 1 in 13 million)

### what is the probability of winning the lottery if choosing 6 numbers from 1 to 30?

Picking 6 numbers out of 30, where you make exactly 6 picks to get those 6 winning numbers, where all 30 possible numbers are different, picked numbers can't be repeated, and the order in which the numbers are picked doesn't matter: Buying 1 ticket: P(first picked number is a winner) = 6/30 P(2nd picked number is a winner) = 5/29 P(3rd picked number is a winner) = 4/28 P(4th picked number is a winner) = 3/27 P(5th picked number is a winner) = 2/26 P(6th picked number is a winner) = 1/25 P(all 6 of the above events happen) = 6/30 * 5/29 * 4/28 * 3/27 * 2/26 * 1/25 P(all 6 of the above events happen) = (6 * 5 * 4 * 3 * 2 * 1) / (30 * 29 * 28 * 27 * 26 * 25) P(all 6 of the above events happen) = 24! * 6! / 30! P(all 6 of the above events happen) = "30 choose 6" P(all 6 of the above events happen) = 30 nCr 6   (on a calculator) P(all 6 of the above events happen) = 0.00000168413  (about 1 in 600,000)   Buying 100 tickets and you want to win at least once: P(at least 1 win) = 1 - (1 - 0.00000168413)^100 P(at least 1 win) = 0.00016839896   (about 1 in 5,938.3)   Buying 100 tickets and you want to win exactly once (not twice or more): P(win exactly once) = P(win on a single draw) * ( P(lose on a single draw) )^99 * (# ways to arrange 1 win and 100 losses) P(win exactly once) = 0.00000168413 * (1 - 0.00000168413)^99 * 100 P(win exactly once) = 0.00016838492  (about 1 in 5,938.8)