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Mathematics 10 Online
OpenStudy (anonymous):

A spinner is divided into 10 equal sections numbered 1 through 10. If the arrow is spun twice, what is the probability the first number will be even and the second number will be a 7? I dont understand how to do this problem

OpenStudy (kropot72):

The probability of the first number being even is: \[P _{1}=\frac{quantity\ of\ even\ numbers}{total\ quantity\ of\ numbers}\] The probability of the second number being a 7 is: \[P _{2}=\frac{1}{10}\] The results of the two spins have no effect on each other, therefore they are independent events. The probability the first number will be even and the second number will be a 7 is given by: \[P _{1} \times P _{2}=you\ can\ calculate\]

OpenStudy (anonymous):

thank you :) helped a lot

OpenStudy (kropot72):

You're welcome :) Have you a result?

OpenStudy (anonymous):

0.1 ?

OpenStudy (kropot72):

Not really. \[P _{1}=\frac{4}{10}\] \[P _{2}=\frac{1}{10}\] \[P _{1} \times P _{2}=\frac{4}{10}\times\frac{1}{10}=can\ you\ calculate?\]

OpenStudy (anonymous):

0.04 but how did you get 4/10 ?

OpenStudy (kropot72):

Look at my first post for how to find the value of P1.

OpenStudy (anonymous):

but the answer is 0.04 or did i get it wrong again?

OpenStudy (driftracer305):

it will be 1/2 and 1/10

OpenStudy (driftracer305):

so 0.5 and the other one will be 0.1

OpenStudy (kropot72):

@driftracer is correct. There are 5 even numbers.

OpenStudy (driftracer305):

yea so sorry i jumped directly to 1/2.............it would be 5/10.......which equals to 1/2.........equals to 0.5....

OpenStudy (anonymous):

In a large bag of candy, each of the 5 colors (red, orange, yellow, blue, and green) occurs with the same probability. You reach in and select 1 candy. Would you use the same equation to this one ?

OpenStudy (kropot72):

\[P _{1}=\frac{5}{10}\] \[P _{2}=\frac{1}{10}\] The probability the first number will be even and the second number will be a 7 is: \[P _{1} \times P _{2}=\frac{5}{10}\times\frac{1}{10}=you\ can\ calculate\]

OpenStudy (anonymous):

Find P(Red, orange, or yellow). sorry i didnt finish the question

OpenStudy (driftracer305):

P(red) =1.........same goes for orange and yellow

OpenStudy (kropot72):

The probability of selecting each of the five colors = 1/5. The probability of selecting Red, orange, or yellow is 3/5

OpenStudy (anonymous):

so would it be (1/5) x (1/5) x (1/5) = 0.6 ?

OpenStudy (kropot72):

3/5 = 0.6 \[\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=\frac{3}{5}=0.6\]

OpenStudy (anonymous):

okay thank you :)

OpenStudy (kropot72):

You're welcome :)

OpenStudy (driftracer305):

ohh wait......u meant all-together.......i thought separately...........sorry man...but yea.............3/5 is thumbs up..

OpenStudy (anonymous):

x A B C D P(x) 0.30 0.20 ? 0.28 Find P(A and B) assuming that A and B are independent. Do you just add up A & C ?

OpenStudy (anonymous):

I Mean C & D *

OpenStudy (anonymous):

@kropot72

OpenStudy (driftracer305):

it will be 0.22

OpenStudy (kropot72):

@jamilexl Can you post the whole question please?

OpenStudy (driftracer305):

he did...as a different new question

OpenStudy (kropot72):

Thanks.

OpenStudy (anonymous):

im a she lol but i dont think that matters

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