# What are the answers and how do you solve them?

May 15, 2018

The odds are 1/10, 3/10, 6/10, in that order. It's a dependent event.

#### Explanation:

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 $\frac{2}{5}$ since there are $2$ blue marbles but $5$ total marbles.

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 $\frac{1}{4}$

$\frac{2}{5} \cdot \frac{1}{4} = \frac{2}{20} = \frac{1}{10}$, so P(blue,blue)=1/10

P(red,red) involves a similar approach. The first time, the odds of picking a red marble is $\frac{3}{5}$ since there are $3$ red marbles, and $5$ marbles total.

If you pick the red marble, there is $2$ red marbles remaining, and $4$ total marbles remaining, making the odds of picking a red marble on the second try equal to $\frac{2}{4}$

$\frac{3}{5} \cdot \frac{2}{4} = \frac{6}{20} = \frac{3}{10}$

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 $\frac{1}{10}$, and the odds of picking two red marbles is $\frac{3}{10}$. The odds of picking two marbles of the same color in a row is $\frac{4}{10}$ then, or $\frac{2}{5}$

Therefore the odds of picking two different marble colors (P(one blue, one red) is $1 - \frac{2}{5} = \frac{3}{5}$

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 $\frac{2}{5}$, and the odds of picking a red marble after that is $\frac{3}{4}$. Therefore the probability of the event taking place is $\frac{2}{5} \cdot \frac{3}{4} = \frac{3}{10}$, meaning P(blue, red; in that order) is $\frac{3}{10}$

So in conclusion

P(blue,blue)=$\frac{1}{10}$
P(red,red)=$\frac{3}{10}$
P(one blue, one red)=$\frac{6}{10}$
P(blue, red; in that order)=$\frac{3}{10}$

May 15, 2018

$\frac{1}{10} , \frac{3}{10} , \frac{3}{10} , \frac{6}{10}$ and dependent.

#### Explanation:

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
$\frac{2}{5} \cdot \frac{1}{4} = \frac{1}{10}$ (B, B)
$\frac{3}{5} \cdot \frac{2}{4} = \frac{3}{10}$ (R,R)
$\frac{2}{5} \cdot \frac{3}{4} = \frac{3}{10}$ (B, R)
Getting one of each is also the sum of the chance of either outcome
$\frac{2}{5} \cdot \frac{3}{10} + \frac{3}{5} \cdot \frac{2}{4} = \frac{6}{10}$ (1B, 1R)
We can check that this is correct as the probabilites of (B, B), (R, R), (B,R) and (R, B) add up to 1