Question

A bag contains 4 red marbles and 6 blue marbles. Brandon selects one marble from the bag. Without replacing the marble he then selects a second marble. What is the probability that he first selects a blue marble and then selects a red marble?

Answer 1

2 out of 5 is the probability

Answer 2

Answer:

The probability that he first selects a blue marble and then selects a red marble = $\frac{4}{15}$

Step-by-step explanation:

A bag contains 4 red marbles and 6 blue marbles.

The total number of marbles = 4 + 6  = 10

Let's take "A" be the event of selecting blue marble.

We know the probability = The number of favorable outcomes / The total number of possible outcomes.

So, the probability that he first select a blue P(A) = 6/10

Now let's take "B" be the event of selecting red marble.

Here the key word is "Without Replacing"

So we will have only 9 marbles, because already one is taken.

Therefore, the total number of possible outcomes = 9

the number of favorable outcome(selecting red marble) = 4

The probability that select a red marble P(B) = 4/9

It is the probability of  dependent events (without replacement)

So, the probability that he first selects a blue marble and then selects a red marble = P(A). P(B)

= $\frac{6}{10} *\frac{4}{9}$

= $\frac{24}{90}$

To simplify, we can divide both the numerator and denominator by 6

the probability that he first selects a blue marble and then selects a red marble = $\frac{4}{15}$