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GED Math : Statistics

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Example Question #41 : Probability

A spinner is divided into five equal quadrants, each of which with a different color: red, blue yellow, green, and orange. If you roll a fair six-sided dice and then spin the spinner, what is the probability of rolling then landing on either red or blue?

Possible Answers:

$\frac{1}{8}$

$\frac{1}{30}$

$\frac{1}{15}$

$\frac{17}{30}$

Correct answer:

$\frac{1}{15}$

Explanation:

Start by finding the individual probabilities of either event.

The probability of rolling a is $\frac{1}{6}$ .

The probability of landing on either red or blue on the spinner is $\frac{2}{5}$ .

To find the probability of both events happening, you must multiply the two probabilities together.

$\text{Probability of rolling 4 AND landing on red or blue}=\frac{1}{6}\times \frac{2}{5}=\frac{2}{30}=\frac{1}{15}$

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Example Question #42 : Probability

In a bag of marbles, are blue and are red. What is the probability of randomly selecting a blue marble?

Possible Answers:

7/13

7/20

13/20

13/7

Correct answer:

13/20

Explanation:

When dealing with probability problems, you must set up a fraction. In the numerator, or the top part of the fraction, you put the number of whatever is being asked for, and in the denominator, or the bottom part of the fraction, you put the total number of items in the sample.

In this case, you are being asked for the probability of selecting a blue marble. Since there are blue marbles, goes in the numerator of your fraction. Since there are blue marbles and red marbles, there are a total of marbles in the bag. So is the total number of items in your sample, and is the number that goes in the denominator of your fraction.

Therefore, your answer is 13/20 .

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Example Question #43 : Probability

In a jar of marbles there are red marbles, white marbles, and blue marbles. What is the probability of drawing a red marble and then drawing a blue marble, without replacement?

Possible Answers:

$\frac{9}{95}$

$\frac{1}{30}$

$\frac{11}{95}$

$\frac{3}{28}$

Correct answer:

$\frac{9}{95}$

Explanation:

First, find the probability of drawing a red marble.

Recall what a probability is: $\frac{\text{Number of desired outcomes}}{\text{Total number of outcomes}}$

Since there are red marbles and a total of , the probability of drawing a red marble is $\frac{12}{20}$ .

Since we are not replacing the marbles, after we've drawn one marble, the total number of marbles left will be .

Next, find the probability of drawing a blue marble. Since there are blue marbles and a total of marbles, the probability of drawing a blue marble is then $\frac{3}{19}$ .

Since the question wants the probability of both events happening, we will need to multiply the two probabilities.

$\text{Probability of drawing red marble then blue}=\frac{12}{20} \times \frac{3}{19}=\frac{9}{95}$

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Example Question #44 : Probability

Out of a standard deck of cards, you draw one card. What is the probability that the card is an ace or black?

Possible Answers:

$\frac{1}{2}$

$\frac{15}{26}$

$\frac{29}{52}$

$\frac{1}{26}$

$\frac{7}{13}$

Correct answer:

$\frac{7}{13}$

Explanation:

The events are NOT mutually exclusive, meaning there is some overlap. That is, there exist cards that are both black AND an ace. Therefore, we have...

P(Ace or Black) = P(Ace) + P(Black) - P(Ace and Black)

In a deck there are 4 aces, 26 black cards, and 2 cards which are both black and an ace.

$P(Ace)=\frac{4}{52}$ P(Black)=26/52 $P(Ace\bigcap Black)=\frac{2}{52}$

P(Ace or Black) = $\frac{4}{52}+\frac{26}{52}-\frac{2}{52}=\frac{28}{52}=\frac{7}{13}$

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Example Question #41 : Statistics

At a barber shop, there is a jar of lollipops. There are 10 grape, 1 orange, and 4 cherry. The barber lets you randomly choose one, put it aside, and then randomly choose another. What is the probability of choosing a grape on the first pick, and an orange on the second pick?

Possible Answers:

$\frac{1}{21}$

$\frac{11}{15}$

$\frac{2}{45}$

$\frac{1}{11}$

$\frac{10}{121}$

Correct answer:

$\frac{1}{21}$

Explanation:

You choose a lollipop, and then put it aside before choosing again. This is probability WITHOUT replacement. The probability of choosing a grape on the first pick is $\frac{10}{15}$ because there are 10 grapes out of 15 total. When the second pick is made, one lollipop has been discarded, so there are now 14 left. The probability of getting an orange lollipop on the second pick is $\frac{1}{14}$ . The probability of the two events occurring is $(\frac{10}{15})(\frac{1}{14})=\frac{10}{210}=\frac{1}{21}$

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Example Question #1861 : Ged Math

If four coins are tossed, what is the probability that two heads will come up?

Possible Answers:

$\frac{1}{4}$

$\frac{3}{8}$

$\frac{1}{8}$

$\frac{1}{16}$

Correct answer:

$\frac{3}{8}$

Explanation:

There are two outcomes when tossing a coin. When tossing 4 coins, there are 2 x 2 x 2 x 2 = 16 possible outcomes by the counting principle. Each outcome is equally likely.

It must be noted, however, that more than one outcome fits the CRITERIA for coming up with two heads. For example, there is THHT, TTHH, HHTT...

The number of outcomes that fit the criteria can be found by taking the number of combinations here. In this case, it is $\binom{4}{2}=6$ .

Therefore, there are 6 favorable outcomes out of 16 possible outcomes. The probability of coming up with two heads is $\frac{6}{16}$ or $\frac{3}{8}$ .

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Example Question #47 : Probability

A coin is weighted so that when it's tossed, the probability of landing on heads is .7 and the probability of landing on tails is .3

If the coin is tossed 10 times, what is the probability of it landing on tails at most 3 times (to the nearest thousandth)?

Possible Answers:

.383

.269

.650

.200

.350

Correct answer:

.650

Explanation:

The idea of an event occurring AT MOST 3 times, means that the event occurring 0 times, 1 time, 2 times, or 3 times fit the criteria. So, one has to consider all of those possibilities.

Using binomial distribution, we can assess the probability for each event occurring.

P(0 tails) = $\binom{10}{0}(.3)^{0}(.7)^{10}$

P(1 tail) = $\binom{10}{1}(.3)^{1}(.7)^{9}$

P(2 tails) = $\binom{10}{2}(.3)^{2}(.7)^{8}$

P(3 tails) = $\binom{10}{3}(.3)^{3}(.7)^{7}$

P(0 tails or 1 tail or 2 tails or 3 tails) = P(0 tails) + P(1 tail) + P(2 tails) + P(3 tails) [since they are mutually exclusive] = .650

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Example Question #41 : Probability

If a regular 6-sided die is tossed, what is the probability of the event, E, rolling a value less than or equal to 5?

Possible Answers:

$\frac{2}{6}$

$\frac{1}{6}$

$\frac{4}{6}$

$\frac{5}{6}$

$\frac{1}{2}$

Correct answer:

$\frac{5}{6}$

Explanation:

Each outcome on a roll of a regular die is an equally-likely outcome.

In the case of equally-likely outcomes, one can find the probability of an event occurring by dividing the number of favorable outcomes by the number of possible outcomes. The favorable outcomes are 1, 2, 3, 4, and 5. The possible outcomes are 1, 2, 3, 4, 5, and 6. There are 5 favorable, and 6 possible. So...

$P(E)=\frac{5}{6}$

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Example Question #49 : Probability

What is the probability of the event, E, rolling a regular 6-sided die and getting an even number AND flipping a coin and landing on tails?

Possible Answers:

$\frac{1}{12}$

$\frac{3}{8}$

$\frac{1}{2}$

$\frac{1}{3}$

$\frac{1}{4}$

Correct answer:

$\frac{1}{4}$

Explanation:

The probabilities of rolling a die and flipping a coin are independent. The probability of rolling a die and getting an even number is $\frac{3}{6}$ or $\frac{1}{2}$ and the probability of flipping a coin and getting tails is $\frac{1}{2}$ .

$P(E)=(\frac{1}{2})(\frac{1}{2})=\frac{1}{4}$

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Example Question #50 : Probability

There are 6 green marbles, 4 blue marbles, and 2 red marbles in a bag, for a total of 12. You choose one marble, and then put it back, and then choose again. What is the probability of selecting a green marble on the first pick and a blue marble on the second pick?

Possible Answers:

$\frac{1}{12}$

$\frac{1}{6}$

$\frac{3}{4}$

$\frac{1}{8}$

$\frac{5}{6}$

Correct answer:

$\frac{1}{6}$

Explanation:

This question asks to select a marble WITH REPLACEMENT. These are independent events. The outcome of one does not affect the outcome of the other. We have P(blue and green) = P(blue) x P(green). The probability of selecting a green marble is $\frac{6}{12}$ or $\frac{1}{2}$ and the probability of selecting a blue marble is $\frac{4}{12}$ or $\frac{1}{3}$ . So, the probability of selecting one green marble and one blue marble is $(\frac{1}{2})(\frac{1}{3})$ = $\frac{1}{6}$

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GED Math : Statistics

Study concepts, example questions & explanations for GED Math

All GED Math Resources

Example Questions

Example Question #41 : Probability

Example Question #42 : Probability

Example Question #43 : Probability

Example Question #44 : Probability

Example Question #41 : Statistics

Example Question #1861 : Ged Math

Example Question #47 : Probability

Example Question #41 : Probability

Example Question #49 : Probability

Example Question #50 : Probability

All GED Math Resources

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