Calculating Probability - Algebra II

There are 4 aces in a deck of 52 playing cards.

After 1 ace is removed, there are 3 aces and 51 cards remaining:

To determined the probability of both events, multiply the probability of drawing each card:

Selecting a specific sequence of marbles from the bag without replacement is a dependent or conditional probabiliy. If you successfully draw a blue marble, it affects the chances of subsequently drawing a gold marble.

In the notation of probability we can call the probability of drawing a blue marble and the probability of drawing a gold marble .

If we replaced the first marble we drew, we would simply multiply times , but since we aren't we want to know or, the probability of B, given A.

So the probability of A and B is equal to:

In words, you have a six in ten chance of drawing a blue marble. If you successfully do that, you then have a four in nine chance of drawing a gold marble. Multiply those odds together, and reduce the fraction and you get a four in fifteen probability.

First, calculate the number of students not planning on attending an in state college.

Either subtract those planning on attending an in state college from the total...

50 - 20 = 30

...or add up all the other categories.

12 + 8 + 10 = 30

Then divide by the total number of students.

30/50 = 60%

Playing a sport, which we can call P(A), and owning a dog, which we can call P(B), are traits that we can assume are independent. The probability of one does not affect the probability of the other.

So, P(A and B) = P(A) * P(B)

Fill in the probabilities given into this equation.

P(A and B) = (0.65) * (0.48) = 0.312 = 31.2%

The probability of finding either a blue car (A) or a green car (B) is written
P(A or B).

This type of probability can be found using the equation:

P(A or B) = P(A) + P(B) - P(A and B)

We need to solve this equation for P(B).

Two of the terms are given in the prompt:

P(A or B) = 0.33

P(A) = 0.22

The third, P(A and B) can be inferred. Since a car can be only one color,
P(A and B) = 0

Thus,

0.33 = 0.22 + P(B) - 0

P(B) = 0.11

The probability of drawing a face card P(A) or a red card P(B) can be written as P(A or B), and be calculated using:

P(A or B) = P(A) + P(B) - P(A and B)

The first two terms on the right side of the equation are fairly straightforward. Since we'll be adding and subtracting fractions, I will keep the denominator the same and won't reduce the fractions until the end.

P(A) = 12/52

P(B) = 26/52 = 1/2

The third, P(A and B), can be calculated using:

P(A and B) = P(A) * P(B)

P(A and B) = 12/52 * 1/2 = 12/104 = 6/52

(Conversely, think there are 12 face cards in a deck of 52, half of which \[6\] are red.)

Back to our original equation!

P(A or B) = 12/52 + 26/52 - 6/52

= 32/52 = 8/13

Multiply these two probabilities together:

Therefore the probability of flipping a tail on a coin and rolling a 3 is .

Binomial probability

The probability of flipping the coin in this particular order is calculated by multiplying the probabilities of each turn by each other. Because we have a "loaded" coin the probabilities are as follows:

.

Then simplify:

.

This fraction is in its simplest form.

To calculate the probability of two independent events occuring, you multiply the probabilities of each event occuring independently.

In this case, the probability of rolling any number is , so the probability of rolling this particular combination is:

.

A die has 6 faces with a number on each face from one to six.

Calculate the total number of combinations of the 2 dice.

Write out the combination sets of the condition that satisfy the scenario.

There are only three sets that will achieve the sum of 2 or 3, but no greater than 3. Divide the total number of possibilities by the total number of combinations.

The probability of rolling a sum that is no greater than 3 is:

There are ten marbles in the bag and there are only two colors. One of the colors is red and there are six of them.

Because it's asking for red, we write a fraction.

The top of the fraction represents what the question is asking and the bottom represents the total possible outcomes.

So our answer is

.

Let's list all possible outcomes.

We have H T, H, H, T T, T H.

There are four possible outcomes so that represents the denominator.

Since we are looking for two tails, we only have one possible outcome which is the numerator.

Our final answer is

.

There are three face cards which are J Q K.

Next, there are four suits for every face card which gives us a total of cards we are looking for.

Finally, since there are cards in a deck, we do a fraction of .

There is no answer given however if we divide top and bottom by we get an answer of

.

Our perect squares are

.

There are ten of them. Since there are slips in the bag, our fraction becomes .

If we cut a zero from both top and bottom, which is the same as dividing by ten, our final answer is

.

In a die, there are six sides numbered

.

Prime numbers are numbers with a factor of one and itself.

The prime numbers are .

We have three outcomes and it's divided over six possible outcomes.

So our fraction is or .

In a standard deck, there are thirteen kind of cards from . There are four suits. To have a pair is to have the same kind of card but with different suit. Since we are looking for any pair, the first card Bob picks is anything so the probability is .

Because he keeps the card, we have cards remaining. To get that same card, there are only of them left in the deck. So the chance of getting a pair becomes .

We don't have this choice but if we divide both sides by we get .

The probability of getting a question wrong is .

Since we want to get it wrong twice, we need to multiply as these are events that must occur.

So we do

.

When we see or, we must add probabilities. Since there are socks and we are interested in blue or yellow, we add those totals up.

There are blue or yellow socks to get.

Our fraction now becomes

.