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Degger [83]
4 years ago
14

A spinner is divided into many sections of equal size. Some sections are red, some are blue, and the remaining are green. The pr

obability of the arrow landing on a section colored red is 9 over 20. The probability of the arrow landing on a section colored blue is 6 over 20. What is the probability of the arrow landing on a green-colored section?.
Mathematics
2 answers:
tangare [24]4 years ago
7 0

Answer: Probability of the arrow landing on a green colored section is \dfrac{1}{4}

Step-by-step explanation:

Since we have given that

Probability of the arrow landing on a section colored blue = \dfrac{6}{20}

Probability of the arrow landing on a section colored red = \dfrac{9}{20}

Since there are only three sections i.e. red, blue and green.

We need to find the probability of the arrow landing on a section colored green.

Required probability is given by

1-(\dfrac{9}{20}+\dfrac{6}{20})\\\\=1-\dfrac{15}{20}\\\\=\dfrac{20-15}{20}\\\\=\dfrac{5}{20}\\\\=\dfrac{1}{4}

Hence, Probability of the arrow landing on a green colored section is \dfrac{1}{4}

choli [55]4 years ago
5 0
The answer would be 5/20 because you add the probability of red and blue (15/20), then subtract from the total (20) which would make it 5/20.
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Answer:

(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 : Result on 1st die.

- Random Variable X_2: Result on 2nd die.

- Random Variable X = X_2 - X_1.

Solution:

 

- Possible outcomes of X : { - 2 , -1 , 0 ,1 , 2 }

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                 ( 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 ):3*P ( X_2 = 1 )*P ( X_1 = 1 )

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               P ( X = 1 ): 2* P ( X_2 = 2 ) * P ( X_1 = 1 )

                                : 2* ( 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 )  

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- The distribution Y = X_2,

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                         P(Y=2) = 1/ 3

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