Question

Can someone please help me with this problem? I don't understand it and I would really appreciate the help. I will medal the person with the best answer. Thanks!

Sally decides to do an experiment to see how many times her MP3 player will shuffle through her songs and play her favorite group, “The Googly Girls” per hour. Her results are shown in the probability distribution table below. Find the probability that the MP3 player will play 1 or 2 songs by “The Googly Girls” each hour.

x 0 1 2 3
p(x) 0.13 0.67 0.14 0.06

0.14
0.19
0.67
0.81

Answer 1

p(x=1 or 2)=p(x=1)+p(x=2)-p(x=1 and x=2)

so
p(x=1 and x=2)=0

so
p(x=1 or 2)=p(x=1)+p(x=2)-0
=0.67 +0.14

Answer 2

So the answer is 0.81?

Answer 3

Thanks amoodarya! Can you help me with one more similar one?

Answer 4

sure

Answer 5

Thanks! Look at the attachment please!

Answer 6

e(x)=p(x1)*x1 +p(x2)*x2 +...

Answer 7

E(x)=1(1/36)+2(1/18)+3(1/18)+4(1/12)+5(1/9)+7(0)+….

Answer 8

So I multiply all of them together? I got 43/36.

Answer 9

Are you still there @amoodarya?

Answer 10

it is greater than 43/36

please show me your work

Answer 11

Okay. All I did is multiply and add the numbers you gave me:

E(x)=1(1/36)+2(1/18)+3(1/18)+4(1/12)+5(1/9)+7(0)+….

Answer 12

What that what I was supposed to do?

Answer 13

E(x)=1/36+2(1/18)+3(1/18)+4(1/12)+5(1/18)+8/18+9/36+10/18+12/9+15/18+16/36+18/18+20/18+24/18+25/18+30/18+36/36

Answer 14

Okay. I did it and came up with 11.65. I guess I did it wrong again.

Answer 15

I did it again and came up with 12.277777778

Answer 16

1/(36) [1+2+6+12+...)
I think it 12

Answer 17

It is 12? okay I guess I did it wrong. Are you pretty sure that that is it?

Answer 18

almost sure

Answer 19

Okay thanks so much. I am taking a geometry course and in the very last section they started adding a bunch of statistics. Not nice. I was really confused by these two problems. Thanks so much for your assistance!

Answer 20

glad 2 help u

Answer 21

:) I might need a little bit more help later today. Would it be alright if your on openstudy if i message or tag you? I don't want to do that if it will bug you though.