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?
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dan do you have i thought on this???????
chance of red = 100% - chance of no red
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
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
use a calculator
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)