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likoan [24]
3 years ago
14

The relative frequency of picking up a red toy is . The relative frequency of picking up a green toy is . It is likely that the

machine has the greatest number of toys. It is likely that the machine has the least number of toys.
Mathematics
1 answer:
Roman55 [17]3 years ago
6 0
What subject is this? ill comment the answer when i know
You might be interested in
We have two fair three-sided dice, indexed by i = 1, 2. Each die has sides labeled 1, 2, and 3. We roll the two dice independent
Bogdan [553]

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 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.

7 0
3 years ago
An insurance company has 10,000 automobile policyholders. The expected yearly claim per policyholder is $240, with a standard de
elena55 [62]

Answer:

almost 0%

Step-by-step explanation:

Given that for an insurance company with 10000 automobile policy holders, the expected yearly claim per policyholder is $240 with a standard deaviation of 800

using normal approximation, the probability that the total yearly claim exceeds $2.7 million is calculated as follows:

Sea sumatoria de x = SUMX, tenemos que:

P (SUMX \geq 2700000) = P(\frac{SUMX - 240*10000}{800 *\sqrt{10000} } \geq \frac{2700000 - 240*10000}{800 *\sqrt{10000} })

= P (z\geq \frac{2700000 - 240*10000}{800 *\sqrt{10000}})

= P (z => 3.75)

= 1 - P ( z < 3.75)

P = 1 - 0.999912

P = 0.000088

Which means that the probability is almost 0%

4 0
3 years ago
If a pizza is cut into 8 pieces and 3 children each eat one piece, what fraction of the pizza is gone
liberstina [14]
The answer is 3/8 
hope it helps
please mark as brainliest i need one more
7 0
4 years ago
Read 2 more answers
Small boxes contain DVDs, and large boxes contain one gaming machine. Three boxes of gaming machines and a box of DVDs weigh 48
Pani-rosa [81]
I don’t know so sorry maybe ask someone else but just go back to the passage and see if you can find it
7 0
3 years ago
Find the greatest common factor. 2x3y + x2y2 + 4xy3
Vika [28.1K]
2x^3y + x^2y^2 + 4xy^3

GCF = xy
3 0
3 years ago
Read 2 more answers
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