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Suppose the probability of snow tomorrow is 0.3 while the probability of [ winning the basketball game tomorTow is 0.6. Assuming these events are independent, what ...

Question

Suppose the probability of snow tomorrow is 0.3 while the probability of [ winning the basketball game tomorTow is 0.6. Assuming these events are independent, what is the probability that it snows and IU loses?0.6(B) 03(C) 0.18(D) 0.42(E)(F) 0.28(G) 0.12(H)

Suppose the probability of snow tomorrow is 0.3 while the probability of [ winning the basketball game tomorTow is 0.6. Assuming these events are independent, what is the probability that it snows and IU loses? 0.6 (B) 03 (C) 0.18 (D) 0.42 (E) (F) 0.28 (G) 0.12 (H)



Answers

The probability that Event $X$ will occur is $0.3 .$ The probability that Event $Y$ will occur is $0.6 .$ Given that Events $X$ and $Y$ are mutually exclusive, what is the probability that Event $X$ or Event $Y$ will occur? A. 0.18 B. 0.2 C. 0.3 D. 0.4 E. 0.9

So basically probability in this case is going to be area of the inside rectangle divided by the area of the bigger rectangle or the backboard. So I have two times 1.5 divided by six times 3.5 and solving that I get 0.14 So the answer is option a.

Here an option. E. We have to find whether the games are the two games are independent, so improbability. The two events are independent. If the incidents off one even does not affected, probably off the other event. So in this case, the probably off winning the second game depends on the first game, so both are not independent, so the games are not. Next. We have to find the probability that you lose both the games. So, using the compliment truth, we have probability you lose. The first game is equal to one minus probability that you end off his game, but so the probability that you when the first game is giving us, you know point for So yeah, one minus 0.4. But is there a 0.6? No, we have to find probability that you lose the second game after losing the first game. Probably that you lose a second game after losing first after is in the fest. No, this is nothing but one minus. Trouble did that to win the second game after losing first. The to win second have to losing first, so probably that you win the second him after losing the first. This 0.3 that is 0.7 now using multiplication Drew. Okay, Okay. We have probability that you lose board the games. Zeke will do. Uh, probably that you lose the first game into probability that you lose a second game after losing the first game. So we have 0.6. You told 0.6 in 20.7. It is their boring for two. That is 42%. No. Next we need to find the Probably that you win board begins. So probably that you win the first game do. In the first game. He's given a 0.4 and probability that you win the second game after winning the first game. Windows second after Winnie first this, uh, 0.2. So probability to win board the games is equal to 0.40 point two. The 0.8 which is 8%. Next we need to find the probably more for eggs. Were X is a random, very book that implies a number off games Children to here. Let X be the number off games you win. So the probably Mordor movie X will take the values that zero. It's not winning any of the game. Winning anyone off the game, winning both the games, the respective probabilities that is losing board the games before it is 42% age, but just zero point for two and probably off winning both the games. Following that, his eight person dish, that is, it'll points and wait. So probably opening anyone off the game will be one minus point for two minus points, their weight, so it will be 0.5. Next. We need to find the expected value, understand a deviation. Now here the expected value. Your fix is given acceleration extent to P O fix that is equal to zero in 20 point photo place one in tow. Zero point fight. Five plus two in tow, 0.0 it. What is he? Quit a little. Place your point. Five place 0.16 That is 0.66 Next, we need to find this town a deviation. First, we'll find a radiance signal square that is equal to summation X minus mu the whole square into P off eggs. That is the first value of fix zero minus mu GIs 0.66 the whole square into 0.42 plus one, minus 0.66 the whole square into 0.5 less tu minus 0.66 The whole square into 0.8 that is equal to it's you is your point 183 nous 0.5 Need bliss. Okay, Do appoint one soulful just equal to 0.38 for six. So signals quiet off point rate for six, which is 0.62

In question 25. We have a probability problem that says you play two games against the same opponent. Their probability you in the first game is 0.4. If you win the first game, the probability you also in the second is point to if you lose the first game, the probability that you in the 2nd 0.3 So looking at a few questions part A are the two games independent Well, independent means that it doesn't matter what happens in the first game, that the probability of second game remains the same. In this case, that's that's not true at the the probability of winning the second greatly depends on the outcome of the first. Yeah, alright, M b. What's the probability that you lose both games, the probability of loss and loss? So we know that the probability of winning the first game is 0.4. Therefore, the probability of losing the first game is 0.6 and means multiply. It says. If you lose the first game, the probability that you win the second is 0.3, which means the probability of losing the second game after you already lose the first game. This point 7.6 times 0.7 this 0.42 See, what's the probability you win both games, So probability of when and win same thought process Here it says The probability of winning the first game is 0.4. I am. It's multiply. And then it says the probability that you also win the second is point to so 0.4 times point to this 0.8 Andy let the random variable x b the number of games you win. So we're going to say X is the number of games you win. Find the probability model for X, So probability model is pretty simple here. What you want to do is lift out all the possible outcomes and the probabilities of those happening so you could win zero games and we figured the probability of one is zero games, 2.42 in Part B. You could win one game. I haven't figured that one out yet, and you could win two games, which is win and win. We did figure out the probabilities 20.8 eso the thing about a probability models that all the probability should add up to 100% or one. So the one figure we're missing, we can easily see that 10.4 to 1.8 make 50%. So the missing piece is 50% as well, because those three probabilities will now add up to 100% on E. What are the expected value and standard deviation to the expanded value? In this case of X is the same thing is the mean of X, which is the summation of all X value. Tom's their probability. So we have zero times 00.42 which is nicely just zero plus one times 2.5 plus two times 20.8 We can add that together. We know that zero times anything is zero. We get point 50 for this 0.16 and are expected. Value is 0.66 which means we're expecting to win less than one game here. The standard deviation of X is thes square root of the summation of every single X value, minus the mean squared times. It's probability. So this would be zero minus 00.66 squared times 0.42 one minus 0.66 squared times 0.50 plus and we'll come down the next line to minus 0.66 squared times. It's probability and all that's under the radical, and that gives us extended deviation of 0.62

So the probability of an event occurring is your 0.9 Well, the full probability, um, is basically that has to add up to one because it can't be greater than or less than one house at upto one, which is equivalent to 100%. So if the probability of something happening is 0.9 I could just say that 0.9 plus the probability of something not happening is equal to one. In which case I isolate for this. All we have to do is attract is your point. You're not from both sides. I do one minus 0.9 which gives us 0.91


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