Question
3) What is the probability that the average number of suffering days of longboard users is less than 10 days?Sclcct ]Select _ 0.1840.145 0.222
3) What is the probability that the average number of suffering days of longboard users is less than 10 days? Sclcct ] Select _ 0.184 0.145 0.222
Answers
No participant threw the same number of bullseyes on two different days. If a participant is selected at random, what is the probability that the selected participant threw 3 bullseyes on Day 1 or Day 2, given that the contestant threw 3 bullseyes on one of the three days?
Okay, so for this question, we've got a random variable X. That's uniform from 20 2 36 and we want to find the expectation. Well, we know the expectation is just the middle. So that's gonna be 28 for a uniformity very like this. Then we want to find the probability that X is greater than 30 days. Well, there are a total of or I guess, let's think of it this way. Uh, from 20 2 36 we have a length of 16 right? From 30 to 60 we have a length of six. So our total probability here is going to be this length 6/16 which is the same as 3/8 or 0.375 So these will be our final answers.
In this problem, we have to study the key. One table and uncertainty cautions. Given Bilow, the first question asked is vorticity probability that vacationer will stay for four days. Rip Able shows Greg 18. Practitioners spare for four days. Who gets the number of big business? We add all the free Qurans is so be right, Kourtney less Pardon me. Bless. It's me. Yes. Bring beating. Plus Leverett. Yes, but be less sinks. This is equal roof. But drink be. This is the total number Off vacationers is but 20. So the proof have anything dead over a petitioner Wednesday for four days his knee west do number of vacation is that stay for four days with this page me divided by the door print number off freakishness for this right brand beat. So this is equal to three derided, right, Bring me. But this the program really that occasionally stay for four days is three divided by 20 which is a little required probability you Now the second caution asked is what is the probability conduct of vacationers stay for less than four days to find the number vacationers that stay for less than four. Place me magnifique grand sees part of ignition nurse. Next, Jake for Brian May. No, one day is not here. I guess at three days that this Kourtney less but me revisit ones would bring me seven. So the little baby sitting next revision stay for less than four days. Is he right by? Is it going to 27? Provided by the door. Bring a bottle. Freakishness this right, Rainsy. So, Chris, Izzy went through? No. You like, paid by part B Breakfast. The programming director. Vacation offense Stay for less than four days. Is nine divided by 30. Bitch is our required probability. Now, the third question asked is what person brought? Vacationers stay for more than six days, your friend. A number of vacationers, but staying for more than six base. The AG roughly grants is for the vintage nurse. Let's stay for seven days at eight babies. Then this would be plus six because he will protect me. Six So percent vacationers who stay for more than six days. Is it going through the number of freakishness Take stay for more than six days. They just took me six divided by the total number of red Ph knows reversed run, frank. Me, my big like by hundreds. Well, Chris is equal to 3 60 You write a bright, very busy quinto that people sing breakfast. The percent of vacationers that stay for more than six days is 13%. Bitch is off required answer.
This question we're told that a website randomly selects among 10 products to discount and the color printer is discounted. So the probability Of the colour printer being discounted is basically .1 and were asked for the expected number of days until this product is. And these countries of the expectation It's basically one over p. which is one over .1 that's equals to 10 part B. Were asked the probability that this product is First discounted exactly 10 days from now. The quality that excessive force to tennis at the quality that it's not discounted in the first nine days and it discounted on the 10th day Just close to 0.03 87 that 0.039 And then in part C. were asked if the product is not discounted for the next five days. What's the probability that it's first discounted 15 days from now? So it's not discounted for the next five days. What's the property that was discounted 15. This is from now is basically Since it has no memory, no memory property. So that's again probably takes because 10 which is .039 and then in party. Whereas the probability that this product is first discounted within three or fewer this so that's apology. That is first discounted on the first day discount on the second day or discounted within the first three days. So that's basically equal two 0.1 plus .9 times .1 plus 0.9 square times 0.1 which is equal to .271