Or Randomly Selected Sit There Is A 3.1 Percent Chance That The Leaving Digits Of The Check Amount

=/A “perfect run”?I’m sorry, but that definition is so loose it’s literally now just a puddle.But, I’ll take what I can get!I vote for Mountain to be my Perfect Run video game!Yep.You play as a floating mountain in the sky.What is the goal?There is no goal.What do you do?Whatever you want.What is the point of it?Exactly. The point is, it’s pointless.There is very little to do. You sit there, a lonely mountain. You wait for a while. Maybe some random piece of Debris will hit you, so you can philosophically debate as to why a chair decided to land atop your snow-capped peak, taking the position of the horse which landed far before it.Or you can decide it’s all just worthless junk, and means nothing.You can play some music if you’d like. There are gramophones that can land on you and play some old-timey music. Or, if you press some keys, you can play musical notes. If you push them enough, you generate a shield, which you can use to protect yourself from things you don’t want hitting you, like Murderous Dwarf Stars or tiny Xyzzyx’s.At any point you can stop, and say to yourself that what you just did was a perfect way to play Mountain, since there is no perfect way, yet any way you play is perfect, because what’s how you play the game.Ok, this got a bit weird but hey, it was fun.Ciao!

Three cards are drawn at random from a well-shuffled deck of cards. What is the probability of drawing an ace, a king, and a jack?Answer: [math]\quad P=\cfrac{12}{52}\cdot\cfrac{8}{51}\cdot\cfrac{4}{50}=\cfrac{384}{132600}=\cfrac{16}{5525}\approx0.29\%[/math]I think that there's a very easy way to visualize this problem.I am making the assumption that a card, once drawn, will not be returned to the deck prior to the next draw. However, if cards are returned, the same logic that I am about to outline would still apply only the second and third probabilities would be different, as we would always be choosing from 52 cards.You have a standard deck of 52 shuffled cards in front of you. You randomly chose one card. There were 52 cards to choose from and there were 4 aces plus 4 kings plus 4 jacks (12 cards) that could turn that would be compatible with the goal. So, the probability of the first draw being successful would be 12 in 52.It doesn't matter whether an ace a king or a jack was drawn on the first draw, your goal will still be maintainable. Just for the sake of argument, suppose that a jack was drawn on the first draw. Now, for the second draw, there would be 4 aces and 4 kings (8 cards) that could turn that would be compatible with the goal and there would be 51 cards to choose from. So, the probability of the second draw being successful would be 8 in 51.Obviously, whether an ace or a king turned on the second draw, on the third and final draw there would be only 4 cards left in the deck the could turn to complete the goal and there would be 50 cards to choose from. So, the probability of the third draw being successful would be 4 in 50.Now, we need only multiply the three probabilities together to get the sought-after probability.

I think I have the answer. We can look at this problem on the following way .First, we have to find the total no: of ways in which the student  can answer the 15 questions . He can answer each question in 4 ways . Therefore , we can find the total no: of ways by using the counting principle and this is 4^15(multiplying the no: of possible ways at each stage).i.e cardinality of the sample space under consideration = 4^15.We have to assume that he selects the options randomly such that probability of selecting any of the 4 answers to each question is same(.25) and that the event of answering each question is independent from the others. In this case, we can use the discrete uniform law. i.e we can find the probability of the event under consideration by counting the no: of outcomes which makes it occur and dividing by the cardinality of the sample space. Now, we only have to find the no: of ways in which he can get exactly 8 questions correct. We can split this process into 2 stages thus. First choose 8 correct questions. This can be done in 15C8 ways. Next, we have to get the other 7 questions wrong. This can be done in 3^7 ways(Again using the counting principle. No: of ways to get 1 question wrong = 3. For 7 questions, this becomes 3^7). Now, we can find the total no: of ways to get exactly 8 questions correct by multiplying the no: of ways at each stage. This is 15C8*3^7.Now , Probability can be found by using discrete uniform law as(15C8*3^7)/4^15=.013.We can check if this answer is correct by finding probabilities for getting exactly 0,1.....15 questions correct and adding them up to see if we get 1 ( didn't do that).

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