Valueisplays the Experimental Probability Distribution.Four Dice Roll Welcome You have 630 points.Play Again? Occurrence #requency of SixesTotal rolls is 100...

You roll a six-sided die 60 times. The table shows the results. For which number is the experimental probability of rolling the number the same as the theoretical probability? (See Example 5.)

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So using our empirical model in number 16 we have We're rolling a die 100 times. So that's their number of trials, and our event here is we get a six 80 times. So 80 out of 100 were asked what was the next. The probability that the next role and in sick or role in six. And that's an estimated probability. So are estimated probability.

Two days a role together, what is the quality that the total score of two days is greater than 10? Right? So for what are the Putin cases each day's head? These numbers, the total sum will be total will be Total cases with me 16-6, which is 36 desired case. Uh huh. Greater than 10. It is asking right? Either 11 or two for 11. The cases are 5, 6 or six weeks, But we're only single case 66. So total is making three set. So we will be just three x 36, Which is one way to do it. Yeah.

Okay, so we're rolling two dice that we assume to be fair, meaning that each outcome from the dice is equally likely to occur. Knowing this, we can just try and count up the total possibilities of outcomes for our two dice. So if our first I says six outcomes and our second ice has six outcomes, the possible combinations to dice rolls will be 36 from here. If we want to figure out the probability that they add up about our two dice rolls at up to six, we just need to figure out how many total ways are dice. Rolls could give us six and divide it by 36. So to get a six on two dice, we could roll a five and a one. We could pull a four and a two three and a three federal, too, and before and we could roll a one and a five. So we're gonna have five outcomes here. So our probability that we roll a six with two dice 5/36. That's the final answer

Okay, so in this problem at Dia thrill 10 times reporter, they want us to find the chance of getting 10 sixes. So the probability of getting 16 is one over six. If we were to rule it 10 times the probability of getting, uh, 10 sixes. So six every time is one of our six, uh, 10 to 1 over 16.6 2/10. There's one over six, 2 10 and that equals one over 6 to 10. If you play that the calculator, it equals 1.65 terms. 10. To make it part B, you want to find the probability of not getting 10 sixes. And since not getting 10 sixes is the opposite of gain 10 sixes within fund this by simply taking the compliment of getting 10 sixes and take the compliment. We just subtract getting tentacles from one that probably did get intensive things from one. So one minus 1.65 times 10 tonight, Bitch, uh, go ahead and put them to cut later and basically equals zero point mine data. Almost one. Basically, what kind of answer for party is basically one, uh, and our final answer for party is 1.65 times tend to native eight and then for part C. They want us to find the probability of all the rules showing five spots or less if the problem related that one look that one rule shows well, they're my response for less is just five over six because, um, fighting the sides show five. Sponsor last only one side shows more than five spots. So if we're all 10 rolls to show by sponsor less the probability of that, it's just gonna be 5 to 6. Tense 5 to 6 and 76 away to attend to power. So that equals about on 0.16 15 says about 16.15% which I find offensive for part C.