A bag contains two red marble and five blue marbles. You draw a marble, record it's color, replace it, and then drawl a second marble. What is the probability that you will draw a blue marble at least once? A) 0.28 B) 0.49 C) 0.71 D) 0.92 @CollateralDamage @dan815

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A bag contains two red marble and five blue marbles. You draw a marble, record it's color, replace it, and then drawl a second marble. What is the probability that you will draw a blue marble at least once? A) 0.28 B) 0.49 C) 0.71 D) 0.92 @CollateralDamage @dan815

Mathematics
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i say c
there r 7 marbles in all, 5 r blue
so 5/7 is 0.71428571428

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|dw:1442628989097:dw|
dan do you have i thought on this???????
chance of red = 100% - chance of no red
|dw:1442629071301:dw|
no red means both times it was blue that chance both was blue is 5/7*5/7 1-(5/7)^2=1-25/49 = 24/49=approx 0.49
shoot i guess i was wrong...
i read half the problem
@dan815 at least one blue need to include 1 red and 1 blue
Thanks guys
oops i was solving the wrong thing
chance of blue = 1-chance of no red
So is 0.49 wrong
triciaal has the right idea using a tree like that and dan815's expression should be `1-(2/7)^2` which leads to the same answer if you used the tree method (add up the probabilities found on each branch)
1-(2/7)^2 = 1-4/49=45/49
What's that as a decimal
use a calculator please
|dw:1442629431663:dw|
use a calculator http://web2.0calc.com/ and type in the fraction to convert to decimal
you should understand the tree diagram
So it's 0.92
?????????
should be same as P(at least 1 blue) = 1- P (all red)
Am I right
|dw:1442629747769:dw|
yes, `45/49 = 0.918367`
yes

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