Since a die is tossed seven times, for a), we want 4 comes up four times and a number other 4 three times. So,
(1/6)^4(5/6)^3
I think you can do the rest now.
Sorry for posting so many topics, but I really need help on this stuff. I recently lost my work I did on a separate sheet from 2 weeks ago, and now I can't seem to find the answer again.
a fair six-sided die is tossed seven times in a row.
A) What is the probability that a "4" comes up exactly four times? (I got 4375/279936)
B) What is the probability that a "4" comes up an even number of times? (I got 2315/4374)
C) What is the probability that a number higher than "4" Comes up exactly four times (I got 280/2187)
Can anyone help me with this. I need to know how I got these answers again.
3 answers
not that simple
suppose we call getting a 4 is success (S) and not getting a 4 if failure (F)
one possibilty is SSSSFFF
The probability of that is (1/6)^4(5/6)^3 which is what you have.
another is SFSFSFS which would also be (1/6)^4(5/6)^3
in other words we have to count the number of arrangements of SSSSFFF which is C(7,4) or 35
So the prob of Tom's event is 35(1/6)^4(5/6)^3 or 4375/279936
which was Tom's answer.
So Tom was correct.
suppose we call getting a 4 is success (S) and not getting a 4 if failure (F)
one possibilty is SSSSFFF
The probability of that is (1/6)^4(5/6)^3 which is what you have.
another is SFSFSFS which would also be (1/6)^4(5/6)^3
in other words we have to count the number of arrangements of SSSSFFF which is C(7,4) or 35
So the prob of Tom's event is 35(1/6)^4(5/6)^3 or 4375/279936
which was Tom's answer.
So Tom was correct.
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