edwardp03418
19.04.2020 •
Mathematics
Derek wants to display 6 of the marbles in his collection in 2 different showcases. If each showcase must have at least 1 marble, how many different ways can he display the marbles?
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Ответ:
6
Step-by-step explanation:
This is a case of combination since the order does not matter, it only matters that there must be a marble in each one inside, therefore we must apply its formula.
xCn = n! / x! * (n - x)!
n, in this case it would be 4 because we must subtract 2 from the total 6, since there must already be one inside in each display case.
x = 2, since it is the number of showcases.
Replacing:
2C4 = 4! / 2! * (4 - 2)! = 6
Which means that you can in six different ways.
Ответ:
(a) P(X = 0) = 1/3
(b) P(X = 1) = 2/9
(c) P(X = −2) = 1/9
(d) P(X = 3) = 0
(a) P(Y = 0) = 0
(b) P(Y = 1) = 1/3
(c) P(Y = 2) = 1/3
Step-by-step explanation:
Given:
- Two 3-sided fair die.
- Random Variable X_1 denotes the number you get for rolling 1st die.
- Random Variable X_2 denotes the number you get for rolling 2nd die.
- Random Variable X = X_2 - X_1.
Solution:
- First we will develop a probability distribution of X such that it is defined by the difference of second and first roll of die.
- Possible outcomes of X : { - 2 , -1 , 0 ,1 , 2 }
- The corresponding probabilities for each outcome are:
( X = -2 ): { X_2 = 1 , X_1 = 3 }
P ( X = -2 ): P ( X_2 = 1 ) * P ( X_1 = 3 )
: ( 1 / 3 ) * ( 1 / 3 )
: ( 1 / 9 )
( X = -1 ): { X_2 = 1 , X_1 = 2 } + { X_2 = 2 , X_1 = 3 }
P ( X = -1 ): P ( X_2 = 1 ) * P ( X_1 = 3 ) + P ( X_2 = 2 ) * P ( X_1 = 3)
: ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 )
: ( 2 / 9 )
( X = 0 ): { X_2 = 1 , X_1 = 1 } + { X_2 = 2 , X_1 = 2 } + { X_2 = 3 , X_1 = 3 }
P ( X = -1 ):P ( X_2 = 1 )*P ( X_1 = 1 )+P( X_2 = 2 )*P ( X_1 = 2)+P( X_2 = 3 )*P ( X_1 = 3)
: ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 )
: ( 3 / 9 ) = ( 1 / 3 )
( X = 1 ): { X_2 = 2 , X_1 = 1 } + { X_2 = 3 , X_1 = 2 }
P ( X = 1 ): P ( X_2 = 2 ) * P ( X_1 = 1 ) + P ( X_2 = 3 ) * P ( X_1 = 2)
: ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 )
: ( 2 / 9 )
( X = 2 ): { X_2 = 1 , X_1 = 3 }
P ( X = 2 ): P ( X_2 = 3 ) * P ( X_1 = 1 )
: ( 1 / 3 ) * ( 1 / 3 )
: ( 1 / 9 )
- The distribution Y = X_2,
P(Y=0) = 0
P(Y=1) = 1/3
P(Y=2) = 1/ 3
- The probability for each number of 3 sided die is same = 1 / 3.