A die is rolled 20 times and the number of twos that come up is tallied. Find the probability of getting exactly four twos. A. 0.202 B. 0.075 C. 0.101 D. 0.083
This is a pretty standard probability question. What is the P of getting 4 twos in a row, followed by 16 rolls with no twos?
And then, what is 20 choose 4?
choose ? x
How many ways are there to choose 4 items from a set of 20. These are called combinations.
how do i make a formilu
I'm getting there. The formula has two parts. The probability in my first post above times the value of what's called 20 choose 4, the number of combinations of 4 items from 20 total. The first term is \[(1/6)^{4} (5/6)^{16}\] The second term is 20 x 19 x 18 x 17 / 4!.
Can you do that multiplication?
20x19=380x18=6840x17=116280/4=29070
Yes, for the second term. How about the first?
Actually, I get a different answer for the second term. Let Python do the work: >>> a = (1.0/6)**4 >>> a 0.0007716049382716048 >>> b = (5.0/6)**16 >>> b 0.05408789293239543 >>> c = 20 * 19 * 18 * 17 / 24.0 >>> c 4845.0 >>> a*b*c 0.2022035812171727
Read the first part of this: http://en.wikipedia.org/wiki/Binomial_distribution
that looks krazy
Well, OK. But to say it again, the probability of 4 twos in a row followed 16 rolls with no twos is: \[(1/6)^{4} (5/6)^{16}\] You agree with that?
yes
but how did you get this >>> a = (1.0/6)**4 >>> a 0.0007716049382716048 >>> b = (5.0/6)**16 >>> b 0.05408789293239543 >>> c = 20 * 19 * 18 * 17 / 24.0 >>> c 4845.0 >>> a*b*c 0.2022035812171727
Maybe I shouldn't have posted that. It's a programming language called Python and I use it to do moderately complex calculations without making mistakes. Notice that I obtained one of your possible answers as my answer.
The trick to the calculation is to realize that although the probability of getting your 4 twos right at the beginning is very low (about 4.17 x 10^-5.. in the problem statement they can come at any point during the series. That's a combinations problem. And the answer they show is the classic formula.
The second "they" being the wikipedia article.
is this one the same If 5 apples in a barrel of 25 apples are rotten, what is the expected number of rotten apples in a sample of 2 apples? A. 0.4 B. 1 C. 0.63 D. 0.33
Anyway, that's how you do this kind of problem. Can you guess for a series of n rolls what is the most likely value for the number of twos? Hint: it's what you would naturally guess.
Umm.. The P of getting a bad one one for the first pick is 1/5, and almost 1/5 for the second one (depending on what happened the first time). The expected value is the sum of these, I think… Anyone else?
The answer is found from the binomial distribution as follows: \[P(4twos)=\left(\begin{matrix}20 \\ 4\end{matrix}\right)(\frac{1}{6})^{4}(1-\frac{5}{6})^{16}=0.2022\]
I think you have a typo
And for @85295james, that funny thing in the parentheses at the beginning of his formula is the official way to say "20 choose 4"
o
You are right. \[P(4twos)=\left(\begin{matrix}20 \\ 4\end{matrix}\right)(\frac{1}{6})^{4}(1-\frac{1}{6})^{16}=0.2022\]
Yes, that's good.
so the p of getin two rotten apples in a row is high but how do i put that in math
\[\left(\begin{matrix}20 \\4\end{matrix}\right)=\frac{20!}{4!16!}\]
Which is the same as 20 x 19 x 18 x 17 / 4 x 3 x 2
Well, for your apples problem, it is just about 1/5 plus 1/5 so I'd go with 0.4.
116280 / 24 =486.66
Do the division again..
this is "/" divide
116280 divided by 24 is equal to 4845
o yes it is
See ya. Think about my bonus problem.
ok
but how about the apples
See above, about 10 posts up.
1/5 + 1/5 =0.4 how did you get that
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