What are the answers and how do you solve them?
The odds are 1/10, 3/10, 6/10, in that order. It's a dependent event.
P(blue,blue) is equal to the odds of picking blue the first time times the odds of picking blue the second time. The first time, the odds of picking blue is
If you pick a blue marble, there is 1 blue marble remaining, and 4 total marbles remaining, making the odds of picking a blue marble on the second try equal to
P(red,red) involves a similar approach. The first time, the odds of picking a red marble is
If you pick the red marble, there is
The last one you can do out step by step, following a similar method above, but there is a shortcut you can use for this case.
The odds of picking two blue marbles is
Therefore the odds of picking two different marble colors (P(one blue, one red) is
There are two possibilities for P(one blue, one red) to happen, one of which is P(blue, red; in that order).
The odds of picking a blue marble first is
So in conclusion
P(one blue, one red)=
P(blue, red; in that order)=
This is a dependent event as the second one is based after which one you took in the first one.
Knowing that, we calculate the chance of getting the first one multiplied by the chance of getting the second one from the remaining ones
Getting one of each is also the sum of the chance of either outcome
We can check that this is correct as the probabilites of (B, B), (R, R), (B,R) and (R, B) add up to 1