A jar contains four red marbles numbered one through four,
and twelve blue marbles numbered one through twelve. A marble is drawn at random from the jar.
Find the probability of the given event. We have four probabilities to find.
Let's begin by sketching the sample space. So we have four red marbles
numbered one through four,
and we have twelve blue marbles numbered one through twelve.
Notice how we have a total of sixteen marbles, four red ones,
and twelve blue ones. And since the probability of an event
is equal to the favorable number of outcomes, divided by the total number of
outcomes.
Since we're selecting just one marble, the total number of outcomes
to reach probability will be sixteen. So first we want to find the probability that
the marble
drawn from the jar is red. Well notice how there are four
favorable outcomes, because there are four eight marbles.
And therefore, the probability of selecting a red marble would be four,
sixteenths, which does simplify, there's a common factor of four.
So this would simplify nicely to one-fourth,
which we should recognize as a decimal
as zero, poinst two, five, which as a percent would be
twenty-five percent. Remember a probability can be expressed
as a fraction, decimal, or percentage. Though most of our homework does asks for a fraction,
or a decimal. Next we want to find the probability
that we'd select an odd marble. Again a total number of outcomes would be sixteen,
because we're selecting one marble
from this group. And now let's count the odd
marbles. Well there are two odd red marbles.
One and three. And there are six
odd blue marbles. One, three,
five, seven, nine
and eleven. So there are a total of eight favorable outcomes
out of sixteen. So the probability of selecting an odd marble is eight sixteenths,
which again simplifies, a common factor of
eight, simplifies to one-half,
which the decimal would be zero point five,
as a percent, fifty percent. Remember, to convert from a decimal to percent,
we multiply by one hundred and add a percent.
Which is the same as moving the decimal point to the right, two places.
And this next probability is probably the most challenging one.
We could use a special formula to find this probability,
since the event of being red, and being odd
are not mutually exclusive. Meaning the marble can be both red and
odd. So we have to be careful not to count any of these marbles twice.
So again, we're selecting one marble, so our total number of outcomes would be sixteen.
Let's count the marbles that are red or odd. Notice there are four
red marbles here. So these are favorable.
We're only going to count them once, so we're not going to count these again,
when we count the odd marbles. So looking for the odd marbles that we haven't counted
here,
there are six odd blue marbles.
One, three, five,
seven, nine, and eleven. So we have a total of four plus six or ten
favorable outcomes. So the probability of being red or odd,
is ten-sixteenths. Simplifying we have a common factor of
two, which gives us
five-eighths as our probability of a marble being
red or odd. We should recognize this as
zero point six, two, five as a decimal. But we can always divide to check.
Notice five divided by eight is
zero point six, two, five. As a percent,
we'd have sixty-two point five percent.
And now for the last probability,
we want to find the probability that the marble is blue
and even. So to be favorable, the marble has to satisfy both of these conditions,
meaning it has to be blue, and it has to be even.
So we're only considering the blue marble's here.
And notice how six of them are even. Two,
four, six, eight, ten,
and twelve. Which gives us six favorable outcomes.
So the probability of a marble being blue and even,
is six-sixteenths. Simplifying,
common factor of two, this gives us
three-eights. Which would be equal to zero point three, seven, five,
or thirty-seven point five percent.
I hope you found this helpful.