Question 1032922: Consider the experiment where a pair of fair dice is thrown.
Let X denote the random variable whose
value is determined by multiplying the number of spots showing on
the one die by the number of spots showing on the other.
The range of values that X can assume are the
positive integers {1,2,3,4,5,6,8,9,10,12,15,16,18,20,24,25,30,36}.
Please give the corresponding probabilities for the values of X given below.
Pr(X = 3) =
Pr(X = 4) =
Pr(X = 5) =
Pr(X = 6) =
Pr(X = 8) =
Pr(X = 9) =
Pr(X = 10) =
Pr(X = 12) =
Pr(X = 15) =
Pr(X = 16) =
Pr(X = 18) =
Pr(X = 20) =
Pr(X = 24) =
Pr(X = 25) =
Pr(X = 30) =
Pr(X = 36) =
Further, find the probability that X is divisible by 15.
Probability that X is divisible by 15 equals.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Consider the experiment where a pair of fair dice is thrown.
Let X denote the random variable whose
value is determined by multiplying the number of spots showing on
the one die by the number of spots showing on the other.
The range of values that X can assume are the
positive integers {1,2,3,4,5,6,8,9,10,12,15,16,18,20,24,25,30,36}.
1.... 1 way
2.... 2 ways
3.... 2 ways
4.... 3 ways
5.... 2 ways
6.... 4 ways
8.... 2 ways
9.... 1 way
10... 2 ways
12....4 ways
etc. Figure these out up to 36 which can be gotten in 1 way
Count the total # of ways the products can be gotten. (Ill call that "n")
Please give the corresponding probabilities for the values of X given below.
Pr(X = 3) = 2/n
Pr(X = 4) = 3/n
Pr(X = 5) = 2/n
Pr(X = 6) = 4/n
Pr(X = 8) = 2/n
Pr(X = 9) = 1/n
etc.
Pr(X = 10) =
Pr(X = 12) =
Pr(X = 15) =
Pr(X = 16) =
Pr(X = 18) =
Pr(X = 20) =
Pr(X = 24) =
Pr(X = 25) =
Pr(X = 30) =
Pr(X = 36) = 1/n
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I'll leave the following to you::
Further, find the probability that X is divisible by 15.
Probability that X is divisible by 15 equals.
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Cheers,
Stan H.
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