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May 22nd, 2011, 12:37 PM  #1 
Senior Member Joined: Sep 2009 Posts: 106 Thanks: 0  [Solved] Probability A or B
hi, I have some more questions. Hire is the problem: Code: A person is about to ride down the mountain 3 times. Probability of falling is 20%. What is the probability that he will fall: 1) Each time? 2) First time, but not fall second & third time? 3) At least once {Solutions: 0,8%; 12,8%; 48,8%} 0 >Not fall 1) There is only 1 favorable outcome [FFF] out of 8 possible, so we must divide 1/8 , why does this not work ? 1.1) Lets try different approach: we need [F & F & F] thus we must multiply 0,2 * 0,2 * 0,2 = 0,008 , again why it wont work? 2) I dont know how to even start this 3) so we need [F or F or F] so [0,2+0,2+0,2] = 0,6, again wrong, why ? These are the rules of probability why dont they work ? 
May 22nd, 2011, 11:14 PM  #2 
Senior Member Joined: Apr 2011 From: USA Posts: 782 Thanks: 1  Re: Probability A or B
I can answer most of this, but I don't know a good way to explain why 1 doesn't work. I think it's simply cause you're not taking the 20% probability into account. As for 1.1, it does work. .008 = .8% 2. What is the probability of not falling? .80. (1  .20) Follow how you did 1.1, except keep in mind that first time is a fall, and the 2nd and 3rd times are not falls and that you have to consider the probability of not falling for those. i.e. .2 x .8 x. 8. We can do this call the fall/not fall have very specific "slots," so it's just a multiplication rule. 3. It says "at least once," not that there is one fall that can be any of the 3 trips down. F + F + F is saying that they fall the first time OR the second time OR the third time. That's not the same thing, cause it only allows for one fall, as opposed to one, two OR three falls. "One, two or three" isn't the same as "one fall on first, second or third" trip down. (And it wouldn't be .2 for each one but that's another story.) You already know there are 8 possible ways to fall/not fall. You'd actually have to figure out the probabilities of each of those and then add together all the ones that are "at least one." FFF is one possibility  and you figure that the way you did 1.1. FF0 is .20x.20x.80 Etc. But 7 of your 8 possibilities have at least one fall so that's a lot of work. You can do it that way though. When doing one way is too long, it's sometimes easier to look at it from the opposite direction. That is, all probabilities must add up to 1. So the 8 of them must add up to 1. So you can figure out the probability of what you don't want and subtract from 1. That is, 000 is not falling at all. All of the other 7 possibilities have at least 1 fall in them. So the probability of one of those 7 choices is going to be 1  P(not falling, or 000). Go back to what I said for 2. The P(not falling) = .8. So P (000) = .8 x .8 x. 8. But that's the one of seven possibilities you don't want. So 1  (.8 x .8 x. is your answer. Just wondering  do you know the binomial equation? That's what this is but you're doing the manual way. 
May 22nd, 2011, 11:31 PM  #3 
Senior Member Joined: Apr 2011 From: USA Posts: 782 Thanks: 1  Re: Probability A or B
Here, I just thought of this. Maybe it'll help. You've probably worked with coin tosses  everyone does. This is the same sort of thing, but it's not on the surface apparent. The reason is that for coin tosses, it's .50 for heads and .50 for tails. The probability for success and failure are exactly the same, so it doesn't matter which way you do it. For 3 coin tosses, your possibilities are: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT 8 total ways, pretty much like your mountain falls. And you probably already know each one has a 1/8 probability of happening. That's how you tried to initially approach the falls, except that falling/not falling aren't .50/.50. So think of the coins this way instead. P(H) = .50 P(T) = .50 So P(HHH) = .50 x .50 x .50. You think of it (probably) as 1/8 because you have 8 possible outcomes. But that only works cause the heads and tails are the same. .50 x .50 x. 50 = .125 = 1/8. P (HHT) = .50 x .50 x .50 = .125 = 1/8. Etc. But the H and T are both .50. So each one of these is going to neatly come out to 1/8 each. But... your fall isn't .50. It's .20. P (fall) = .20 and P (not fall) = .80. So now let's look at it the same way we did the coins: FFF, FF0, F0F, F00, 0FF, 0F0, 00F, 000 But now let's use multiplication and apply the .20 and .80 to it: P (FFF) = .20 x .20 x. 20 = .008 (which does not equal 1/8, because the F and 0 aren't equally weighted like coins) P (FF0) = .20 x .20 x .80 = .032 (which also does not equal 1/ Compare this to the coins. They don't come out the same cause they aren't both .50. One is .20 and one is .80. But the concept is still there. The 7 I bolded up there are your "at least one" choices. The long way is using this method to calculate the probability of each one of those 7, and add them together. But the 1  P(000) is easier. (Am I just confusing you more? Sorry if so.) 
May 23rd, 2011, 01:45 AM  #4 
Member Joined: May 2011 Posts: 51 Thanks: 0  Re: Probability A or B
^^^ Erimess, your 2nd time approach is superb. Most of the time when there are only 2 outcomes people tend to think as 5050 and that is the trick here. BTW 'If you don't need a data given in the question to derive the answer, then your approach is wrong' I just wish I could find a shorter way for this, how about riding down 50 times? 
May 23rd, 2011, 11:47 AM  #5  
Senior Member Joined: Sep 2009 Posts: 106 Thanks: 0  Re: Probability A or B Quote:
I will need to go & digest this now, thank you for your answer.  
May 23rd, 2011, 10:05 PM  #6 
Senior Member Joined: Apr 2011 From: USA Posts: 782 Thanks: 1  Re: [Solved] Probability A or B
Yes, it's a lot to digest. The binomial looks like this. If it doesn't look familiar then don't worry about it until (if) you get to it. 

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