SOLUTION: (Binomial Distribution)
three normal six-sided dice are thrown.find the probability of obtaining
(a) three odd numbers.
my answer:
X~Bin(18, 1/2) <-----because i think ther
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Probability-and-statistics
-> SOLUTION: (Binomial Distribution)
three normal six-sided dice are thrown.find the probability of obtaining
(a) three odd numbers.
my answer:
X~Bin(18, 1/2) <-----because i think ther
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Question 387336: (Binomial Distribution)
three normal six-sided dice are thrown.find the probability of obtaining
(a) three odd numbers.
my answer:
X~Bin(18, 1/2) <-----because i think there will be 18 numbers
<-----(1/2) will be odd numbers and another (1/2)will be even no.
P(X=3) = [(18C3) * {(1/2)^3} * {(1- 1/2)^15} ]
= 51/16384
but,i think my answer is wrong.please correct me if i'm wrong. Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! (Binomial Distribution)
three normal six-sided dice are thrown.find the probability of obtaining
(a) three odd numbers.
-----------------
# of ways to succeed.
There are 3 odd numbers on each die.
# of ways to get three odd #'s = 3^3 = 27
----
# of random outcomes = 6^3 = 216
----
P(3 odd numbers) = 27/216 = 1/8
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Cheers,
Stan H.
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