Question 21 ptsImagine throw a fair seven-sided die once: Which of the following does not belong to the event space of this experiment?{1,2,3,4,5,6,7}{3,6, 12}{3 , ...

A die is thrown repeatedly untill a six comes up. What is the sample space for this experiment?

So find the probability that when it waas a die twice, you will get a some off. What? The probability that you get a some off to is some off three? All a Some off, off, off. Mhm we have already. We did this table from the previous question, so we're not going to repeat it again. But we just have to identify from the table where we get the some off too. So we'll get the summer off to here some of three and the sum of 12. So they are 12 three for such outcomes. So therefore, the probability is able to for over 36 cities are simple space. You're history.

Okay, Head in this case are die is tossed toys simultaneously. We just need to find the probability of die, Vince, which is? The sum is equals to six. So probability that some of numbers in those two days is six rate. So some is six. We need to figure out those cases out of 36 cases. So let's say that even this a so easy equivalent one cases went comma by because it really six. The second case will be too common for the third case with three common three. The fourth case will, before coming to the fifth case, will be five cummerbund, right? And if you just see this three commentary if we just interchange, this will remain three comma three. So this will not count ice. So there are a total of six cases. So favorable number of pieces. It's six. We have Putin number element in the total number of element in the sample space equivalent to tactics or end off SS equals to 36 which is six times six. So instead required probability that some principles to six it's 6/36 6/36 is when over six

All right. So we are tossing a single die And a single die has six faces. Um What is the probability and the odds of getting a fall? So we start with the probability of getting and four and this will be equal to How many forces are on the faces of the Dye one divided by the sample space, which is successes. And to find the odds, odds of getting a four. We are going to work out. Uh the probability of the favorable outcome of getting a four already calculated as a ratio of the probability of not getting the favorable outcome, which is Then six or 5. Remember it is one of the six divided by five or 6. But when you put a multiplication with the need to invent and this will give us 1/5 the second one in the same way we have done the first part. What is the probability of getting a six? It's still won six out of the symbol space? How about the odds of getting a six Probability of a successful outcome? It's 1/6 not applied by which which is not flying by. What would they have divided with? So this would be 6/5 and this would give us one of our five. Like the first depart now the third part. What is the probability of getting a four or a six? So probability or for all six? So this would be equal to look at it. We have got how many um successful outcomes, It's a four or a six. So there are two Out of the symbol space which is six. And certainly if you reduce this would be 1/3. Now what are the odds of getting a four or a 6? So firstly we always start with the successful outcome Uh getting a 46 which have already been calculated and it's one of the three. If I did buy To over three within the probability or the complement of the successful outcome. So if he would divide, if you multiply, it would then invade to 3/1, No. And this will give us 1/2, and that is uh uh the answer.

Three days a toast. And we have to find the probability that is six turns upon exactly one day. So we have three possibilities. That first one is the first day shows six and others are not six. Second Witness. Second show six others and about six. And a third witness. Third day of the show, six others to at North six. So first even oneness. First to -6 others had not six. Any others. Similarly. Second minus faster. Any North sixth, 2nd 1 is six and 3rd one is not six and third monday's 1st, 10 2nd. They're not six And Hard -6. So first let's find the probability for the first event. Even so, probably to get six on first days one by six and not six on first days five or six in five by six. That is equal to 25 x 216. Similarly, for the second possibility, it's also five x 6 in the one x 6 Into five x 6. Do you think about it too, 25 divided by 2, 1, 6, and for the third possibility also, It is equal to 25 Bye two and 6. Therefore, the total probability is Equal to 25 plus 25 plus 25 divided by 2 16. That is equal to 75 x 216. This is the required probability.