### Video Transcript

A jar of marbles contains four blue
marbles, five red marbles, one green marble, and two black marbles. A marble is chosen at random from
the jar. After replacing it, a second marble
is chosen. Find the probability that the first
is blue and the second is red.

One of the key parts to this
question is the fact that the marble is replaced. This means that we are dealing with
independent events. The first marble does not impact
the selection of the second marble. This is because the total number of
marbles in the jar will remain constant. Every time that a marble is
selected from the jar, there will be a total of 12 marbles to choose from. We know that when dealing with
independent events, the probability of event A and event B occurring is equal to the
probability of A multiplied by the probability of B. This is known as the
intersection.

In this question, we will let event
A be the probability of selecting a blue marble. Event B is the probability of
selecting a red marble. We can write probability as a
fraction, where our numerator is the number of successful outcomes and the
denominator is the number of possible outcomes for any random event. In this case, the top number or
numerator will be the number of marbles of the color we want, and the denominator
will be the total number of marbles. There are four blue marbles. Therefore, the probability of event
A is four out of 12 or four twelfths. There are five red marbles. Therefore, the probability of event
B, selecting a red marble, is five out of 12 or five twelfths.

Before multiplying these fractions,
we notice that the first fraction can be canceled. Both the numerator and denominator
are divisible by four, so four twelfths simplifies to one-third. We can then multiply the numerators
and, separately, the denominators. One multiplied by five is equal to
five, and three multiplied by 12 is 36. The probability that the first
marble selected is blue and the second is red is five out of 36.