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anonymous
 one year ago
The table shows the results of rolling a number cube, with sides labeled 1 through 6, several times.
What is the experimental probability of rolling a 2 or a 4?
Express your answer in simplest form.
Outcome 1 2 3 4 5 6
Number of times outcome occurred 10 6 4 8 6 6
anonymous
 one year ago
The table shows the results of rolling a number cube, with sides labeled 1 through 6, several times. What is the experimental probability of rolling a 2 or a 4? Express your answer in simplest form. Outcome 1 2 3 4 5 6 Number of times outcome occurred 10 6 4 8 6 6

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@peachpi @Mahmood786123

Mahmood786123
 one year ago
Best ResponseYou've already chosen the best response.0just add all the values up! A 1 was rolled 10 times, a 2 was rolled 6 times...how many "times" were there total?

Mahmood786123
 one year ago
Best ResponseYou've already chosen the best response.0Where's the 94 coming from? I get 40! When you're calculating probability, you want to find how often your outcome of interest occurs out of the total number of outcomes that actually occur. I'm trying to get you to calculate each individually, starting with the total number of outcomes. The number cube was rolled a bunch of times. Each roll is called a trial. The number that was rolled was recorded for each trial performed. In the end, 10 trials rolled a 1, 6 trials rolled a 2, and so on through 6. The total number of trials can be found by adding all the trials that resulted in each outcome, so: 10+6+4+8+6+6=40 There were 40 trials/rolls in total. Of those 40 rolls, we want to know how often a 3 or a 6 popped up. According to the table, 4 rolls resulted in a 3, and 6 rolls resulted in a 6, so out of the 40 rolls, our outcome of interest (a 3 or a 6) occurred 4+6=10 times. That means the experimental probability (the probability based on this particular experiment) of rolling a 3 or a 6 is 10/40, or 1/4. Does that makes sense?

Mahmood786123
 one year ago
Best ResponseYou've already chosen the best response.0It's just a matter of understanding the situation you're given. For these kinds of questions, always try to follow the procedure I explained above!

Mahmood786123
 one year ago
Best ResponseYou've already chosen the best response.0hope I helped :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0u think u can help with 5 more

Mahmood786123
 one year ago
Best ResponseYou've already chosen the best response.0i'll try but I might need to leave soon

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok imma close this then reopen
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