GED Math - Probability | Practice Hub

Given the data set $\left { 2, 3, 5, 6, 6, 6, 7, 7\right }$ , which of the following quantities are equal to each other/one another?

I: The mean

II: The median

III: The mode

II and III only

I and II only

I and III only

I, II, and III

Explanation

Since this data set is arranged in ascending order and has an even number of elements, the median of the data set is the arithmetic mean of its middle two elements. Both elements are 6, so this is the median.

6 is the mode, since it occurs most frequently.

The mean is the sum of the elements divided by the number of elements, which is 8:

$\frac{2+3+ 5 + 6 + 6 + 6 + 7 +7}{8} =\frac{42}{8 } = 5.25$

The median and the mode are equal to each other, but not to the mean, so the correct answer is "II and III only".

A parking lot contains the following:

11 black cars
8 red cars
3 blue cars

Find the probability that the next car that leaves the parking lot is a black car.

$\frac{1}{2}$

$\frac{1}{11}$

$\frac{11}{12}$

$\frac{1}{4}$

Explanation

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

$\text{probability of event} = \frac{\text{number of ways event can happen}}{\text{total number of possible outcomes}}$

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

\text{number of ways event can happen} = 11

because there are 11 black cars in the lot.

We can also calculate the following:

\text{total number of possible outcomes} = 22

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

11 black cars
8 red cars
3 blue cars

Now, we can substitute. We get

$\text{probability a black car leaves} = \frac{11}{22}$

$\text{probability a black car leaves} = \frac{1}{2}$

Therefore, the probability that a black car leaves the parking lot is $\frac{1}{2}$ .

A die is altered so that it comes up a "6" with probability $\frac{1}{4}$ . The other five outcomes are equally likely. If this die and a fair die are rolled, what is the probability that the outcome will be a total of "2"?

$\frac{1}{40}$

$\frac{1}{36}$

$\frac{9}{400}$

$\frac{3}{20}$

Explanation

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 $\frac{1}{6}$ .

In the loaded die, since a "6" will come up with probability $\frac{1}{4}$ , for the other five rolls to be equally likely, each, including "1", must come up with probability

$\frac{1}{5} \left ( 1 - \frac{1}{4}\right ) = \frac{1}{5} \times \frac{3}{4} = \frac{3}{20}$ .

A double "1" will appear with probability

$\frac{1}{6} \times \frac{3}{20} = \frac{1}{2} \times \frac{1}{20 } = \frac{1}{40}$ .

A parking lot contains the following:

7 blue cars
1 white car
3 black cars

Find the probability the next car that leaves is a blue car.

$\frac{7}{11}$

$\frac{1}{7}$

$\frac{7}{4}$

$\frac{1}{11}$

Explanation

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

$\text{probability of event} = \frac{\text{number of ways event can happen}}{\text{total number of possible outcomes}}$

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

\text{number of ways event can happen} = 7

because there are 7 blue cars in the lot

We can also calculate the following:

\text{total number of possible outcomes} = 11

because there are 11 total cars in the lot that could potentially leave:

7 blue cars
1 white car
3 black cars

Now, we can substitute. We get

$\text{probability a blue car leaves} = \frac{7}{11}$

Therefore, the probability that a blue car leaves the parking lot is $\frac{7}{11}$ .

A weighted coin is flipped times. It is weighted such that heads comes up out of times on average. What is the probability that the person in question will flip all tails?

$\frac{16}{625}$

$\frac{8}{5}$

$\frac{2}{5}$

$\frac{8}{25}$

$\frac{81}{625}$

Explanation

If the weighting is $\frac{3}{5}$ for heads, this means that it is $\frac{2}{5}$ for tails. Therefore, the probability of flipping tails will be:

$\frac{2}{5}*\frac{2}{5}*\frac{2}{5}*\frac{2}{5}=\frac{16}{625}$

Billy flips a fair coin three times. His results were three tails. Billy then flipped the coin three times again and was shocked to see the results were also three tails! What is the probability that Billy flipped the coin 6 times, and all trials are tails?

$\frac{1}{64}$

$\frac{1}{3}$

$\frac{1}{12}$

$\frac{1}{128}$

$\frac{1}{4}$

Explanation

Every result for a fair coin is independent of the previous trials.

The probability of landing heads or tails is $\frac{1}{2}$ .

For the coin to land on tails six consecutive times,

$(\frac{1}{2})^6 =(\frac{1}{2})(\frac{1}{2})(\frac{1}{2})(\frac{1}{2})(\frac{1}{2})(\frac{1}{2})$

The answer is: $\frac{1}{64}$

What is the probability of selecting a king from a deck of cards, putting it aside, and then selecting an ace?

$\frac{4}{663}$

$\frac{1}{169}$

$\frac{2}{13}$

$\frac{1}{2652}$

$\frac{1}{2704}$

Explanation

This question involves probability WITHOUT replacement.

In a regular deck of cards, there are 52 cards, with 4 kings. The probability of selecting a king is $\frac{4}{52}$ . Once that card is out of the deck, there are 51 cards left. There are 4 aces out of 51 cards. The probability of selecting an ace out of 51 cards is $\frac{4}{51}$ .

The probability of selecting a kind AND an ace is $(\frac{4}{52})(\frac{4}{51})=\frac{16}{2652}=\frac{4}{663}$

When rolling a $\small 6$ -sided die, what is the probability of rolling $\small 3$ or greater?

$\small \frac{2}{3}$

$\small \frac{1}{2}$

$\small \frac{1}{6}$

$\small \frac{1}{3}$

Explanation

When rolling a die, the following outcomes are possible:

$\small 1,2,3,4,5,6$

Of the $\small 6$ outcomes, $\small 4$ outcomes are $\small 3$ or greater. Therefore,

$\small P=\frac{4}{6}=\frac{2}{3}$

Six boys and four girls are finalists in a drawing. The ten names are placed into a hat, and two are drawn at random. What is the probability that both names drawn will be girls?

$\frac{2}{15}$