How do you solve these probability word problems???

Question

A jar contains 4 blue marbles and 2 red marbles. A marble is chosen at random and is NOT replaced. A second marble is chosen. Find each probability.

1. P(both marbles are red)
2. P(one marble is red AND one is blue)

Please help, I have a problem with the dependent events part of these problems so please show work so i understand the problem instead of just writing some random answer.

Answer 1

For these problems, you multiply the individual probabilities together. By not replacing marbles, it will change what you are picking on the second pick, as well as the total in the jar.

1. P(First marble is red) = 2 red /6 total = 2/6
P(Second marble is red) = 1 red/5 total = 1/5
P(Both marbles are red) = 2/6 * 1/5 = 2/30 = 1/15

2. P(First marble is red) = 2 red /6 total = 2/6
P(Second marble is blue) = 4 blue/5 total = 4/5
P(First red, second blue) = 2/6 * 4/5 = 8/30 = 4/15

Now you could also pick a blue marble, and then a red marble, since the question just says "one red and one blue" and doesn't specify which order. So you also consider this case:

P(First marble is blue) = 4 blue/6 total = 4/6
P(Second marble is red) = 2 red/5 total = 2/5
P(First blue, second red) = 4/6 * 2/5 = 8/30 = 4/15

Now you add these two probabilities together to get 4/15 + 4/15 = 8/15

hope this helps!

Answer 2

A good check is to work out P(both marbles are blue) and make sure that the sum of the three probabilities is exactly 1. There are no other possible outcomes :)

The rule is: Probabilty of A and B = Probability of A x Probability of B given A has already occurred


Getting two red marbles:

p(first one is red) = 2/6
p(second one is red, given first one was red) = 1/5

p(first two are red) = 2/6 x 1/5 = 2/30 = 1/15

The two different ways of getting one red and one blue marble:

p(first one is blue) = 4/6
p(second one is red, given first one was blue) = 2/5
p(first is blue and second is read) = 4/6 x 2/5 = 8/30 = 4/15

p(first one is red) = 2/6
p(second one is blue given first one was red) = 4/5
p(first is blue and second is read) = 2/6 x 4/5 = 8 / 30 = 4/15

p(one red and one blue) = 4/15 + 4/15 = 8/15


Getting two blue marbles:

p(first one is blue) = 4/6
p(second one is blue, given first one was blue) = 3/5
p(first two are blue) = 4/6 x 3/5 = 12 / 30 = 2/5

Total = (2 + 12 + 8 + 8 ) / 30 = 30 / 30 = 1 Correct!

Answer 3

4+2=6 marbles

1) 2 red-1 red =1 red
6-1=5 marbles
P=2/6 * 1/5
=1/15

2) First Red then Blue
P=2/6*4/5
=4/15