Hey guys, just a quick question about probabilities...I'm horrible with them. I am using the destroyer from august 1, 2007. IT's number 64 of the math section. The question is : The weatherman tells you that there's a 70% chance of rain each day for the next 3 days. What's the probability it will rain 2 out of the next 3 days?
The solution says to use the formula:
P = (nCr)(p)^n(1-p)^r
n being the total # trials
r being the # of trials with the desired result
p being the probability of the event occuring
THe book then says this is the solution:
P = (3C2)(.7)^2(.3)^1
???
I am completely lost as to why .7 is ^2 and .3 is ^1
I thought it would be .7^3 (n=3, as shown in the representation of nCr-->3C2) and I thought .3^2 (r=2 as also shown in nCr)
What am I doing wrong here...or is the book wrong?
Also, how do you know when to use nCr, nPr, or the above equation when dealing with other probabilities? Are there any word clues that you can use to help you distinguish which one to use?
Thanks in advance for any and all help with this, I really appreciate it.🙂
Formula is nCr * p^r * (1-p)^(n-r), I don't think anyone has gotten it right yet.
Okay piece by piece here.
nCr first: This is n!/(n-r)!r!
The n! on top would be the # of events while the r is the # of desired successes. Here we have n = 3 days with r = 2 days of rain. Considering n! and (n-r)! alone would be useless. If we look at them together, we notice that n!/(n-r)! actually leaves us with
👎(n-1)(n-2)...(n-r+1). Take a closer look and you'll notice that there are 'r' number of terms in there.
For example if you had n = 10 and r = 4, you would have 10!/(10-4)! = 10!/6! = 10*9*8*7. There are 4 terms there. With r = 4 you want 4 successes so you consider the first 4 terms.
In our example we have n = 3 and r = 2 so that's 3! / (3-2)! = 3!/1! = 3*2. This represents the first two terms of 3!.
nCr asks the question: how many ways can we select 'r' objects from a group of 'n' objects without considering the order we select them in?
Now if we want a specific order that's nPr and that's just n!/(n-r)! which we've already done. To get nCr we throw in that r! in the denominator. Since we are picking 'r' number of successes, how do we account for the number of ways we could have selected them? Given a group of 'r' objects, how many ways could we order them? We could order them in r! different ways.
For example with A B C D E we could pick it in that order, we could do B A C D E, we could do C E D B A, and so forth. If order doesn't matter then we only need to consider ONE of those. That's why we divide by r!. With 2 objects we could have A B or we could have B A. So we divide by 2! = 2. Now we'll only consider 1/2 of them. With 3 objects we could have ABC, ACB, BAC, BCA, CAB, or CBA. So we divide by 3! = 6. Now we'll only consider 1/6 of them.
So that's the nCr (the hardest part of this).
Next is the p^r. The 'p' here stands for probability of course and it typically denotes the probability of SUCCESS. In this case we want rain which has probability of 0.7. The r is the number of successful events. We want 2 days of rain. So we do 0.7^2.
Last is the (1-p)^(n-r). The (1-p) represents probability of FAILURE. Here we have a 0.3 chance of no rain. The (n-r) is the number of non-successful events. We want 1 day without rain. So we do 0.3^1.
Okay so now we group it together.
If you are still confused about why we do 0.7^2 * 0.3^1 then think of it this way:
What if I just wanted to know the probability of rain on day 1, rain on day 2, and no rain on day 3?
You would do 0.7 * 0.7 * 0.3 right? This is 0.7^2 * 0.3^1 which is what we have. That's essentially what that part of the problem does. It takes one example of an outcome that matches what we're looking for. Alternatively you could have chosen rain on day 1, no rain on day 2, and rain on day 3, to get 0.7 * 0.3 * 0.7 which is still 0.7^2 * 0.3^1. It doesn't matter. They are both acceptable outcomes of 2 days rain, 1 day no rain.
The nCr part tells you how many ways this outcome could occur. It's not enough to say 0.7^2 * 0.3^1 because you could have rain, rain, no rain... you could have rain, no rain, rain... you could have no rain, rain, rain. The 0.7^2 * 0.3^1 only accounts for 1 of those. To get the right answer you need the nCr to account for ALL of those.
In total you have 3!/(3-2)!2! * 0.7^2 * 0.3^1
which is 3 * (7/10)^2 * (3/10)
which is (3*49*3) / (1000)
which is 441/1000.
Hope this helps.
