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Algebra 21 Online
OpenStudy (anonymous):

http://s296.photobucket.com/user/ontoonex99/media/Screen%20Shot%202015-06-02%20at%207.37.12%20PM_zpseia5o480.png.html I need help number 18 and 20 please check number 18 did P(A and B)=P(A) X P(B) P(A and B) / P(B)= P(A) = 0.15/0.6 = 1/4 ?

OpenStudy (phi):

I would say \[ P(B|A)= \frac{P(A \cap B)}{P(A)} \] so \[ P(A)= \frac{P(A \cap B)}{P(B|A)} \] with P(B|A)= 0.6 and P(A \(\cap\) B) =0.15 from which you get P(A)= 0.25

OpenStudy (anonymous):

ok,Thanks, can you do number 20?

OpenStudy (phi):

For Q20 there are 20 marbles , 9 of them red. so 9/20 chance of choosing a red marble on the first pick. If you put the marble back in, then you will have 9/20 chance of picking red on the second pick, and 9/20 on the third. thus \[ \left(\frac{9}{20}\right)^3 \]

OpenStudy (phi):

If you do not replace the marble, then the chance of picking a red marble on the first pick is 9/20, on the second 8/19 and on the third 7/18 thus \[ \left( \frac{9}{20}\right)\left( \frac{8}{19}\right)\left( \frac{7}{18}\right) \]

OpenStudy (phi):

for part A, each time you pick is an independent event (because we put the marble back in the bag). That means each pick should be 9/20 and for multiple independent events, you multiply the individual probability to find the probability of all the events occurring.

OpenStudy (anonymous):

thanks , can you have me some more ?

OpenStudy (phi):

ok

OpenStudy (anonymous):

this one I no idea

OpenStudy (phi):

for event A, you count up the number of tiles that are consonants (the problem is y is sometimes a vowel e.g. fly, by, dry, or as a consonant e.g. you, yell, young, ...) not sure how to handle that.

OpenStudy (phi):

and of course for B, first count the number of tiles that are vowels.

OpenStudy (anonymous):

let me know if later you can solve , I need help if you can check my work it correct

OpenStudy (anonymous):

can you check my answer ?

OpenStudy (phi):

Let's assume y is only a consonant. then there are 42 vowels, 2 blanks, 56 consonants the chance of choosing a consonant on the first draw is 56/100 and the chance of choosing a vowel on the 2nd draw is 42/99

OpenStudy (phi):

and we multiply (56/100)*(42/99)

OpenStudy (anonymous):

http://s296.photobucket.com/user/ontoonex99/media/Screen%20Shot%202015-06-02%20at%205.56.17%20PM_zpstctblc9x.png.html \[\frac{ 1 }{6 }\] \[\frac{ 1 }{ 6 } . of .60 = 10 ?\]

OpenStudy (phi):

yes, and the 4 spot came up 10 times

OpenStudy (phi):

yes, so the black marble.

OpenStudy (anonymous):

thanks,next http://i296.photobucket.com/albums/mm168/ontoonex99/Screen%20Shot%202015-06-02%20at%205.56.08%20PM_zpszlf9z9b0.png \[\frac{ 179 }{ 2237 }=0.0800\] 0.0800 x 100= 8% ?

OpenStudy (phi):

I would leave the answer as 0.08

OpenStudy (phi):

yes

OpenStudy (phi):

your picture of the weather does not match up with 9/12000

OpenStudy (anonymous):

let me check

OpenStudy (phi):

for the weather question I think they want you to order A , B, C , D from least to most likely

OpenStudy (phi):

9/1200 (not 12000) and expected number of defects is 9/1200 * 15000

OpenStudy (anonymous):

\[\frac{ 9 }{ 1200 }\times1500?\]

OpenStudy (phi):

yes

OpenStudy (anonymous):

let go back to number 22 D<B<A<C ?

OpenStudy (phi):

chance of rain on Fri is 5%, so chance of no rain is 95% i.e. D has prob= 95%

OpenStudy (phi):

and they want the order smallest to biggest. D will be the biggest chance.

OpenStudy (phi):

what numbers do you get for A, B , C and D ? for A and C, read off the number from the chart for B and D, do 100 - what is on the chart.

OpenStudy (phi):

I get A 80, B 70, C 90, D 95

OpenStudy (anonymous):

thanks you . Take me so long ti fix out , http://s296.photobucket.com/user/ontoonex99/media/Screen%20Shot%202015-06-02%20at%207.35.09%20PM_zpse2zctvy9.png.html 3) Dependent 4) independent

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