Thanks a million!
Hoe willekeurige waarde te genereren op basis van toegewezen waarschijnlijkheid in Excel?
Als er een tabel is met enkele waarden en bijbehorende toegewezen percentages, zoals onderstaand screenshot in een blad. En nu wil ik willekeurige waarden genereren op basis van de lijst met waarden en hun toegewezen waarschijnlijkheden.
Om willekeurige waarden met de waarschijnlijkheid te genereren, hebt u in feite slechts twee formules nodig.
1. Typ deze formule in de aangrenzende cel van de tabel = SOM ($ B $ 2: B2)en sleep deze formule naar de cellen die je nodig hebt. Zie screenshot:
2. Selecteer een lege cel waarin u de willekeurige waarde wilt plaatsen, typ deze formule =INDEX(A$2:A$8,COUNTIF(C$2:C$8,"<="&RAND())+1), druk op enter-toets. En druk op F9 toets om de waarde naar behoefte te vernieuwen.
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- To post as a guest, your comment is unpublished.· 2 months agoYou're a life saver.
Thanks a million!
- To post as a guest, your comment is unpublished.· 4 months agoMÌNH KHÔNG HIỂU LẮM. TẠI SAO PHẢI LÀM NHƯ VẬY Ạ?
- To post as a guest, your comment is unpublished.· 5 months agoOttimo lavoro! Questo è proprio quello che cercavo. Però non funziona con Calc di Open Office, che fra l'altro accetta solo i comandi in italiano.E' possibile tradurla in modo che funzioni anche con calc? Io non ci sono riuscito.Grazie.
- To post as a guest, your comment is unpublished.· 1 years agoTHE FORMULA WORKS ONLY IF ONE SUBSTITUTES RAND() WITH RANDBETWEEN(1;X) WHERE X IS THE TOTAL NUMBER OF POSSIBLE OUTCOMES, WHICH WOULD SIMPLY BY AN INTEGER GREATER THAN 1.
- To post as a guest, your comment is unpublished.· 1 years agofrankly I don't think this formula works.
I am trying to sample the results of an election so I have in a row the votes of different parties. no I want to go in & pick up randomly ten ballots (sort of like exit polling works), but I always get the first party in the rows.
- To post as a guest, your comment is unpublished.· 2 years agoHow can I utilize this random number generator but only have it generate odd or even numbers inside of the criteria? When I add the "ODD" or "EVEN" coding, it only produces the number "1" in the cell.
- To post as a guest, your comment is unpublished.· 2 years agoHi, Preston Ehrsam, the Insert Random Data tool cannot insert random even or odd numbers only.
- To post as a guest, your comment is unpublished.· 3 years agoSorry Just Noticed that I extend the cumulative prob. outside the desired limits
- To post as a guest, your comment is unpublished.· 3 years agoDoes this work if the data in the cells are horizontal instead of vertical? Mine is not. Thanks
- To post as a guest, your comment is unpublished.· 3 years agoHello, can anyone help me getting the correct formula? I would like to get random values but with certain limits. For example, randomly get "red, blue, green, orange or pink", but I have a specific quantity of items for each color so I have to set this condition within the formula.
- To post as a guest, your comment is unpublished.· 4 years agoWhy do the values from this function change every time I make an adjustment to the worksheet? I can't test my sample data if the results keep changing
- To post as a guest, your comment is unpublished.· 1 years agoITS A COMMON ISSUE IN RANDOMLY GENERATED INTEGERS. WHEN THEY ARE GENERATED, THE RANDOM NUMBERS, TAKE THEM & COPY & PASTE SPECIAL THEM AS VALUES, THIS WAY THEY WONT CHANGE ANYLONGER. BUT YOU WILL LOSE THE FORMULA. BEST TO USE A DIFFERENT WORKSHEET IN THE SAME SPREADSHEET.
- To post as a guest, your comment is unpublished.· 4 years agoSorry, the formula will keep changing while adjusting the sheet because the formula contains RAND() function.
- To post as a guest, your comment is unpublished.· 4 years agoAnyone know how I would go about doing this without replacement using probability pools. Here's an example: Pool A has 10 people. Pool B has 7 People. There is an 80% probability we will pull someone from Pool A; 20% probability we will pull someone from pool B. We want to randomly pick 5 people from the 2 pools without replacement. Even without replacement, the probability of pulling from either pool stays at 80/20. Any Ideas?
- To post as a guest, your comment is unpublished.· 4 years agoSorry that I have not found a solustion for your question. If you find a answer, would you mind to let me know？
- To post as a guest, your comment is unpublished.· 4 years ago=INDEX(A$2:A$8,COUNTIF(C$2:C$8,"<="&RAND())+1)
What does the +1 in this formula do?
- To post as a guest, your comment is unpublished.· 4 years agoHello, beacause the cumulative column I count the rows that are less or equal to 1, but the RAND() function only work from 0-0.9999, if you do not +1, the last one value G cannot be randomly choosed forever.