In the first case, since you are returning the marble, the two events are independent, that is,
whatever happened in the first pick has no effect on the second pick
so prob(your event) = (7/16)(7/16) = ....
In the 2nd case you are not returning the marble, so for the second pick there would only be 1 5 marbles left
so .....
let me know what your answer is
there are 4 red marbles 7 blue and 5 green
what is the probability of picking a red replacing it and then getting a blue
what is the probability of picking a red this time setting it aside and getting a blue
why or why not are the answers the same
I have been stuck for a while please help I want to know the way to solve this considering I have mcas soon
4 answers
1. because it would be 4/16 for red and 7/16 for blue I don't know how to make them both into a single probability
2.then there is 4/16 red and 7/15 blue but again I have no clue on how to make them into one
2.then there is 4/16 red and 7/15 blue but again I have no clue on how to make them into one
You just multiply them.
#1 prob = (4/16)(7/16) = 7/184 or appr .038
#2 prob = (4/16)(6/15) = ...
#1 prob = (4/16)(7/16) = 7/184 or appr .038
#2 prob = (4/16)(6/15) = ...
7/60 i think would be the answer for 2
and how did you get 184
and how did you get 184