ACT Math - How to find the probability of an outcome

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:

$\frac{1}{3}$

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

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:

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

A fair die is rolled three times. What are the odds that the die will come up a six on all three rolls?

$\frac{1}{18}$

$\frac{1}{216}$

$\frac{1}{9}$

$\frac{1}{36}$

Explanation

The probability the die will turn up six on any given roll is $\frac{1}{6}$ . To find the probability of multiple events occurring, we multiply the individual probabilities together. $(\frac{1}{6})(\frac{1}{6})(\frac{1}{6})=\frac{1}{216}$

What is the probability that Jim flips a fair coin 3 times and gets 3 heads?

$\frac{1}{2}$

$\frac{1}{4}$

$\frac{1}{8}$

Explanation

The probability of getting heads on one flip of a coin is 1/2, since the only other outcome is tails. Because the flips are independent of each other (one outcome does not depend on the others), the probability of the second coin being heads is 1/2 and the probability of the third coin being heads is 1/2. However, to calculate the probability of obtaining all of the flips ending in heads, we multiply the probabilities of the events together: (1/2)*(1/2)*(1/2) = 1/8.

Kaleah was born on the 17th day of the month. What is the probability that the next person she meets was born the 17th day of the month if they were born in August?

$\frac{1}{31}$

$\frac{1}{30}$

$\frac{17}{31}$

$\frac{1}{365}$

$\frac{2}{31}$

Explanation

To find the probability of an event, find the number of ways that even can happen divided by the total number of possible outcomes.

There are 31 days in August, thus the total number of outcomes for someone being born in August is 31.

Since we only care about the event that is being born on the 17th, there is only one way that event can happen (being born on August 17th).

Thus the probability is

$\frac{1}{31}$ .

A coin is tossed five times. What is the probability of tails occuring at least once?

0.729

1/2

0.833

0.969

3/4

Explanation

Probability of getting a head: 0.5
Probability of getting a tail: 0.5

The probability of not getting ANY tails = 0.5 * 0.5 * 0.5 * 0.5 * 0.5= 0.03125

The probability of getting AT LEAST 1 tail = 1 - 0.03125 = 0.969

If there are 490 marbles in a bag, divided evenly into the seven colors of the rainbow, what is the probability of picking out a marble that is one of the three primary colors?

$\frac{4}{7}$