Probability - GED Math

When rolling a die, the following outcomes are possible:

Of the outcomes, outcomes are or greater. Therefore,

After the removal of two red cards from a deck of 52, there are now fifty cards, of which twenty-four are red. The probability of a random card being red is therefore , which as a percent is

A standard deck of play cards has cards. There are jacks, of which are red. Therefore, the probability is:

For a "2" to be rolled with two dice, both dice must show a "1". In the fair die, this is one of six equally likely outcomes, so it will happen with probability .

In the loaded die, since a "6" will come up with probability , for the other five rolls to be equally likely, each, including "1", must come up with probability

.

A double "1" will appear with probability

.

For a "2" to be rolled with two dice, both dice must show a "1".

For each die, since a "6" will come up with probability , for the other five rolls to be equally likely, each, including "1", must come up with probability

.

The probability of a double "1" showing up will be

.

There are 4 possibilities that a 3 or higher can be rolled on a die.

The probability of rolling a 3 or higher on a fair die is:

Rolling a heads on a coin has a probability of one half. Therefore,

For a roll of "11" to occur with two dice, one die must show a "5" and the other must show a "6". For each die, since a "6" will come up with probability , for the other five rolls to be equally likely, each, including "5", must come up with probability

.

The probability of rolling a red "5" and a blue "6" will be

which is also the probability of rolling a blue "5" and a red "6". Add:

For a roll of "11" to occur with two dice, one die must show a "5" and the other must show a "6". In the fair die, "5" and "6" are two of six equally likely outcomes, so each will happen with probability .

In the loaded die, since a "6" will come up with probability , for the other five rolls to be equally likely, each, including "5", must come up with probability

.

The probability of rolling a "5" on the loaded die and a "6" on the fair die is

.

The probability of rolling a "6" on the loaded die and a "5" on the fair die is

.

Add these probabilities:

The easiest way to determine the probability of at least one head is to determine the complement of this event, which is the probability of two tails. The penny will come up tails 35% of the time (0.35) and the nickel will come up tails 40% of the time (0.40). These are independent events, so the probability of both happening is the product, or

.

The probability of at least one head - the complement of this event - is this probability subtracted from one, so

, or 86%.

A fair dice has 6 faces. Therefore, there is a sixth chance to land on any number between 1 through 6. Rolling the die the first time is independent from the second time, which means that we must account for the probabilities for each trial.

Probability of rolling a 6 is: 1/6

Since probability adds up to 1, the odds of rolling anything other than a 6 for one trial is:

1 - 1/6 = 5/6

On two trials:

Since order is irrelevant here, we are dealing with combinations.

There are

ways to choose two out of ten students.

There are

ways to choose two out of four girls.

The probability that two girls will be chosen is therefore

.

Since order is irrelevant here, we are dealing with combinations.

There are

ways to choose two out of ten students.

There are ways to choose one boy and one girl.

The probability that one boy and one girl will be chosen is therefore

.

To find the probability of an event, we will use the following formula:

Given the event of drawing a red card from a deck of cards, we can calculate the following:

because there are 26 red cards in a deck

• 13 hearts
• 13 diamonds

Now, we can calculate the following:

because there are 52 total cards we could potentially draw.

So, we can substitute. We get

Therefore, the probability of drawing a red card from a deck of cards is .

To find the probability of an event, we will use the following formula:

Now, given the event of a blue car leaves next, we can calculate the following:

because there are 3 blue cars in the lot.

We can also calculate the following:

because there are 9 total cars in the lot that could potentially leave:

• 3 blue cars
• 2 white cars
• 4 black cars

Now, we can substitute. We get

To find the probability of an event, we will use the following formula:

So, given the event of drawing a black card from a deck of cards, we can calculate the following:

because there are 26 black cards in a deck

• 13 clubs
• 13 spades

Now, we can calculate the following:

because there are 52 total cards in a deck we could potentially draw. So, we can substitute. We get

Therefore, the probability of drawing a black card from a deck of cards is .

To find the probability of an event, we will use the following formula:

Now, given the event of selecting a history book, we can calculate the following:

because there are 3 history books on the shelf.

We can also calculate the following:

because there are 18 total books on the shelf we could potentially choose from:

• 10 chapter books
• 5 non-fiction books
• 3 history books

Now, we can substitute. We get

Therefore, the probability of selecting a history book is .

To find the probability of an event, we will use the following formula:

Now, we will look at finding the probability of drawing a King, Queen, or Jack from a deck of cards. We will consider these as 3 separate events. We will find the probability of each, then add them together.

So, we will find the probability of drawing a King from a deck of cards. We get

because there are 4 Kings in a deck of cards:

• King of Hearts
• King of Diamonds
• King of Clubs
• King of Spades

We also get

because there are 52 cards in a deck that we would potentially draw.

So, we get

Now, we will do the same for finding the probability of drawing a Queen from a deck of cards.

because there are 4 Queens in a deck of cards:

• Queen of Hearts
• Queen of Diamonds
• Queen of Clubs
• Queen of Spades

We also get

because there are 52 cards in a deck that we would potentially draw.

So, we get

And lastly, we will find the probability of drawing a Jack. We get

because there are 4 Jacks in a deck of cards:

• Jack of Hearts
• Jack of Diamonds
• Jack of Clubs
• Jack of Spades

We also get

because there are 52 cards in a deck that we would potentially draw.

So, we get

Now, to find the probability of drawing a King, Queen, or Jack, we will add them all together. We get

Therefore, the probability of drawing a King, Queen, or Jack from a deck of cards is .

The deck is divided into four suits: Clubs, Diamonds, Hearts, and Spades. There are 13 of each suit in a standard deck.

The probability of picking out a club out of a plain deck is:

The student does not replace the card back into the deck, which means that there are 51 cards remaining, and there are only 12 clubs remaining in the deck.

The probability of picking out another club would be:

To identity the probability, simply multiply the first and second probabilities.

The answer is:

To find the probability of an event, we will use the following formula:

Now, given the event of grabbing a shirt from the closet, we can calculate the following:

because there are 10 shirts in the closet.

We can also calculate the following:

because there are 15 total things in the closet we could potentially grab:

• 10 shirts
• 2 jackets
• 3 pairs of jeans

Now, we can substitute. We get

Therefore, the probability of grabbing a shirt from the closet is .