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In survey, 159 of the participants said that they had never bought lottery tickets or treasury bonds, 73% had bought lottery tickets and 49% had bought treasury bon...

Question

In survey, 159 of the participants said that they had never bought lottery tickets or treasury bonds, 73% had bought lottery tickets and 49% had bought treasury bonds(a)State the events that describe the scenario given to you:(b)Find the probability that a person chosen at random from those taking ~ part in the survey had bought lottery tickets or treasury bonds (ii) had bought lottery tickets and treasury bonds had bought lottery tickets only

In survey, 159 of the participants said that they had never bought lottery tickets or treasury bonds, 73% had bought lottery tickets and 49% had bought treasury bonds (a) State the events that describe the scenario given to you: (b) Find the probability that a person chosen at random from those taking ~ part in the survey had bought lottery tickets or treasury bonds (ii) had bought lottery tickets and treasury bonds had bought lottery tickets only



Answers

Online Gambling According to a Casino FYI survey, $6.4 \%$ of U.S. adults admitted to having spent money gambling online. If a U.S. adult is selected at random, what is the probability that he or she has never spent any money gambling online?

An online survey discovered six and 4/10 percent of adults admitted to spending money online gambling. If an adult is selected at random, what is the probability that they have not spent money on gambling? Was 100% chance that they either did or didn't? We're certain that one of those two things happened. So now we're going to deal with the compliment, which means we're going to do 100%. Subtract the percentage we have. And if we get rid of the percent that they did spend money and we'll be left with the percent they did not. Obviously you can use a calculator. If we subtract this correctly. It looks like we're gonna have 93 6/10 percent, which is equivalent to 936 thousands. So the probability that an adult is selected randomly and has not spent money and gambling is 93 6/10 percent

Still, Loder is designed so that I pledge, is six members from 1 30 on one but ticket. Ah, what is the probability that a player with one ticket we will What is the probability off 100 different together to win. So you can, uh, so we have to plan the simple space. So they're saying it's from Want to take it in there to truth six. Order the leaders of Major as long as they called the six numbers. Therefore, we use combinations. So it's 30. Cease its en la The probability off. Ah, well, if you a chest wide ticket is it goes to one over simple space you for Bill it off. You win. If you were 100 tickets is it was a one off us a simple space.

For this problem, we are told that selecting lottery numbers or the this problem, we are told that in one version of a popular lottery game, a player select six of the numbers or yeah, six of the numbers from 1 to 50 for the agency in charge of the lottery. Also select six numbers. Then asked what is the probability that the player will match the six numbers? If 2 50 cent tickets, This jackpot is worth at least $2 million dollars in prize money and grows according to the number of tickets soul. So uh, the probability of matching the six numbers, if 2 50 cents or 50 cent tickets are picked are purchased. Well, let's start with considering that We have the probability of getting one number, right? So probability of say one number. So I'll just say one end Is going to be equal to one and 54. So 1/54. Then the probability of getting a the second number. Right. So probability of two in it's going to be that probability of getting the first number right Times the probability of getting the second number right now. Since we've gotten the first number right on our second choice, we aren't choosing amongst 54 instead. We're choosing amongst 53. So if we work through that logic, the probability then of getting six correct? It's going to be won over 54 times 1/53 Times 1/52 times 1/51, Times 1/50 Times let's see here, that's at five now. So one lastly 1/49. But no actually I'll calculate that out first here one moment. So that is going to be a very, very small number which I'm going to just currently express as can write this as being 48 factorial over 54 factorial. Which is why we have that 50. That factorial up until we reach the point of 49, then everything gets divided out. But I'll note that's if we have two tickets, it's alright p of six. And with two tickets Each one of those, we actually have two opportunities to get it correct. So rather than being 48 factorial over 54 factorial, it will be two to the power of six times 40 factorial over 54 factorial. And that is going to come out to a much more meaningful number though still very small. That's going to be about four in Let's see here. Yeah. four in 1,162,222,000 425. So not great odds

Culpability off a person within the price that doesn't know what it be. Off e which is equal toe 666 divided by 53 c six Because we need to to the six number out of six and there are a total numbers are 53 we need to to six number out of 53. So the probability is number of favorable cases divided by the total number of cases. So our probe lt is equal to one a bone 53 multiplied with 52 multiplied with 51 multiplied with 50 multiplied with 49 multiplied, with 48 divided by the six factorial. So which is equal to theta six Factorial divided by 53 multiplied with 52 my deployed with 51 multiplied with 50 multiplied with 49 my deployed with 48 which is equal to 0.0 zero zero 4355 Now probe lt off a person know what winning the prize is equal to one minus probe lt off winning the prize which is equal to one minus 0.0 for 355 which is equal to 0.9999 nine Approximate So this is our answer


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