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mathdummy4
PLEASE HELP a dice is rolled 8 times. find the probability. p(getting at least 7 even numbers) a. 1/256 b. 9/256 c. 1/32 d. 1/16
total number of permutations when a dice is rolled 8 times = 6^8
even numbers : 2, 4, 6 so, out of 8 times u perform experiment, 7 times - u have 3 choices : 3^7 one time you have 6 choices : 6 total permutations in favor = 3^7 * 6
how did you get 3/48??
At least 7 even is the same as at most 1 odd. 0 odd = 1/256 1 odd = 8/256 Total = 9/256
p(getting at least 7 even numbers) = \(\frac{3^7*6}{6^8}\) = \(\frac{3^7}{6^7}\) = \(\frac{3^7}{3^7.2^7}\) =\(\frac{1}{2^7}\) =\(\frac{1}{256}\)
there are 3 even numbers on a dice and 3 odd and you dividethe 3 even by 8 times and that will give you the answer
6times 8 is 48 and divide by 3 and you will get 16 so the answer is 1/16
okay...so what if you were looking for 8 odd numbers? wouldn't it be the same process?
well i got the same answer..1/16...is that right?
Binomial distribution: P(n,i)=(n,i)p^i q^(n-i) (n,i)=n!/(i!(n-i)!) Here, p=q=1/2 (8,0)=1 (8,1)=8 (8,2)=8*9/2=36, ... So any number of even or odd can be calculated.
You have to have 7 throws even and the 8th one can be either. \[ \frac 1{2^7} 8=\frac 1 {16} \]
You multiply by 8 to count where the even or odd throw is.
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