A dies is thrown six times. (a) what is the probability that the nth throw is n on each occasion? (b) what is the probability that the nth throw is n on exactly five occasions?
the first answer is (1/6)^6...
i am not quite sure of the 2nd.. bu i think its (1/6)^5*(5/6).. Still you can confirm with someone..
Wait, I misread the question.
can you help me out with part (b)
How many sides does the dice have? n?
Or just 6?
just 6
So what do you mean by "the nth throw is n"?
the number on the die is the same number as the throw
Oh, ok.
like first throw you get 1. i think thats what it means lol
So for every throw there is a probability of 1/6 of getting the right number.
Thomas9 i believe the first one is correct.. they are independent events.. everytime the probability is 1/6 so you just multiply the probabilities according to multiplication principle
yeah
first one is indeed correct.
the first one is right but the second one isnt
oh...
i have the answer its 5/7776
but im not sure how to get it
It's binomial distributed, so you need to fill in the pdf. http://www.wolframalpha.com/input/?i=binomial+distribution+pdf
n is the number of expirements, it's 6 in this case because we throw 6 times.
p is the probability of succes, 1/6 in this case.
and x is number of times of succes, in this case 5.
\[P(X=5)=\left(\begin{matrix}6 \\ 5\end{matrix}\right)\frac{1}{6}^5\frac{5}{6}^{6-5}=6\frac{1}{6}^5\frac{5}{6}=\frac{5}{6^5}\]
I hope that made sense to you.
ok wait lol is there a multiplication inbetween each one?
yes
\[P(X=5)=\left(\begin{matrix}6 \\ 5\end{matrix}\right)\cdot\frac{1}{6}^5\cdot\frac{5}{6}^{6-5}=6\cdot\frac{1}{6}^5\cdot\frac{5}{6}=\frac{5}{6^5}\]
k thanks heaps:)
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