Data Analysis

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Questions 1 - 10

There are 20 balls in a box, 10 are green, 5 are blue, and 5 are red. One red and one green ball are then removed from the box.

What is the probability that the third ball removed from the box will be green?

$\frac{9}{20}$

$\frac{8}{20}$

$\frac{1}{2}$

$\frac{10}{18}$

Not enough information given

Explanation

This is a probability question, in the beginning, the odds of pulling a green ball were 10/20, which reduces down to 1/2. When 2 balls are removed, the total is now 18, and because there are only 9 green balls, we have a 9/18 chance of pulling a green ball, and this reduces down to 1/2

There are 20 balls in a box, 10 are green, 5 are blue, and 5 are red. One red and one green ball are then removed from the box.

What is the probability that the third ball removed from the box will be green?

$\frac{9}{20}$

$\frac{8}{20}$

$\frac{1}{2}$

$\frac{10}{18}$

Not enough information given

Explanation

When rolling a $\textup{6-sided}$ die, what is the probability you roll a number less than or equal to ? Give your answer as a reduced fraction.

$\frac{1}{3}$

$\frac{1}{2}$

$\frac{1}{6}$

$\frac{2}{6}$

$\frac{3}{6}$

Explanation

To find the probability of an event, find out how many ways that specific event can happen and divide it by the total number of possible outcomes. The only numbers less than or equal to on a sided die are and . There are possible outcomes. Thus the reduced fraction is:

$\frac{1}{3}$

A coin is flipped seven times. What is the probability of getting heads six or fewer times?

$\frac{127}{128}$

$\frac{255}{256}$

$\frac{1}{128}$

$\frac{1}{7}$

$\frac{6}{7}$

Explanation

Since this problem deals with a probability with two potential outcomes, it is a binomial distribution, and so the probability of an event is given as:

$P(n,k)=\frac{n!}{(n-k)!k!}p^k(1-p)^{n-k}$

Where is the number of events, is the number of "successes" (in this case, a "heads" outcome), and is the probability of success (in this case, fifty percent).

One approach is to calculate the probability of flipping no heads, one head, two heads, etc., all the way to six heads, and adding those probabilities together, but that would be time consuming. Rather, calculate the probability of flipping seven heads. The complement to that would then be the sum of all other flip probabilities, which is what the problem calls for:

$P(7,7)=\frac{7!}{(7-7)!7!}(\frac{1}{2})^7(1-\frac{1}{2})^{7-7}=(\frac{1}{2})^7=\frac{1}{128}$

Therefore, the probability of six or fewer heads is:

$1-\frac{1}{128}=\frac{127}{128}$

Presented with a deck of fifty-two cards (no jokers), what is the probability of drawing either a face card or a spade?

$\frac{11}{26}$

$\frac{25}{52}$

$\frac{156}{2704}$

$\frac{3}{52}$

$\frac{9}{26}$

Explanation

A face card constitutes a Jack, Queen, or King, and there are twelve in a deck, so the probability of drawing a face card is $P(F)=\frac{12}{52}$ .

There are thirteen spades in the deck, so the probability of drawing a spade is $P(S)=\frac{13}{52}$ .

Keep in mind that there are also three cards that fit into both categories: the Jack, Queen, and King of Spades; the probability of drawing one is $P(F\bigcap S)=\frac{3}{52}$

Thus the probability of drawing a face card or a spade is:

$P(F)+P(S)-P(F\bigcap S)=\frac{12+13-3}{52}=\frac{11}{26}$

A coin is flipped four times. What is the probability of getting heads at least three times?

$\frac{5}{16}$

$\frac{1}{16}$

$\frac{1}{4}$

$\frac{11}{16}$

$\frac{3}{4}$

Explanation

Since this problem deals with a probability with two potential outcomes, it is a binomial distribution, and so the probability of an event is given as:

$P(n,k)=\frac{n!}{(n-k)!k!}p^k(1-p)^{n-k}$

Where is the number of events, is the number of "successes" (in this case, a "heads" outcome), and is the probability of success (in this case, fifty percent).

Per the question, we're looking for the probability of at least three heads; three head flips or four head flips would satisfy this:

$P(4,3)=\frac{4!}{(4-3)!3!}(\frac{1}{2})^{3}(\frac{1}{2})^{4-3}=4(\frac{1}{2})^{4}=\frac{1}{4}$

$P(4,4)=\frac{4!}{(4-4)!4!}(\frac{1}{2})^{4}(\frac{1}{2})^{4-4}=(\frac{1}{2})^{4}=\frac{1}{16}$

Thus the probability of three or more flips is:

$\frac{1}{4}+\frac{1}{16}=\frac{5}{16}$

Angela scores 17, 19, 13, 24, and 14 points in the first five games of a seven-game basketball season. If the scoring leader in Angela’s league averages 18 points per game, how many points must Angela score in the final two games combined to end the season with the highest scoring average in the league AND have a higher scoring average than any other player?

Explanation

Since a given player’s scoring average can be determined by dividing the sum total of points scored by the number of games, we can determine the total points of the scoring leader by multiplying the average points per game by the total number of games. 18 x 7 = 126. Angela would have to score 1 more point than the current scoring leader. Angela’s current total is 87 points; therefore, she must score 40 (87 + 40 = 127) over the course of the final two games to have the highest average points per game in the league.

When rolling a $\textup{6-sided}$ die, what is the probability you roll a number less than or equal to ? Give your answer as a reduced fraction.

$\frac{1}{3}$

$\frac{1}{2}$

$\frac{1}{6}$

$\frac{2}{6}$

$\frac{3}{6}$

Explanation

$\frac{1}{3}$

A coin is flipped four times. What is the probability of getting heads at least three times?

$\frac{5}{16}$

$\frac{1}{16}$

$\frac{1}{4}$

$\frac{11}{16}$

$\frac{3}{4}$

Explanation

Since this problem deals with a probability with two potential outcomes, it is a binomial distribution, and so the probability of an event is given as:

$P(n,k)=\frac{n!}{(n-k)!k!}p^k(1-p)^{n-k}$

Where is the number of events, is the number of "successes" (in this case, a "heads" outcome), and is the probability of success (in this case, fifty percent).

Per the question, we're looking for the probability of at least three heads; three head flips or four head flips would satisfy this:

$P(4,3)=\frac{4!}{(4-3)!3!}(\frac{1}{2})^{3}(\frac{1}{2})^{4-3}=4(\frac{1}{2})^{4}=\frac{1}{4}$

$P(4,4)=\frac{4!}{(4-4)!4!}(\frac{1}{2})^{4}(\frac{1}{2})^{4-4}=(\frac{1}{2})^{4}=\frac{1}{16}$

Thus the probability of three or more flips is:

$\frac{1}{4}+\frac{1}{16}=\frac{5}{16}$

Presented with a deck of fifty-two cards (no jokers), what is the probability of drawing either a face card or a spade?

$\frac{11}{26}$

$\frac{25}{52}$

$\frac{156}{2704}$

$\frac{3}{52}$

$\frac{9}{26}$

Explanation

A face card constitutes a Jack, Queen, or King, and there are twelve in a deck, so the probability of drawing a face card is $P(F)=\frac{12}{52}$ .

There are thirteen spades in the deck, so the probability of drawing a spade is $P(S)=\frac{13}{52}$ .

Keep in mind that there are also three cards that fit into both categories: the Jack, Queen, and King of Spades; the probability of drawing one is $P(F\bigcap S)=\frac{3}{52}$

Thus the probability of drawing a face card or a spade is:

$P(F)+P(S)-P(F\bigcap S)=\frac{12+13-3}{52}=\frac{11}{26}$

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