Algebra II - Calculating Probability

In a bag has slips of paper each numbered from . What's the probability you pick a perfect square?

$\frac{1}{10}$

$\frac{9}{100}$

$\frac{1}{25}$

$\frac{4}{25}$

$\frac{11}{100}$

Explanation

Our perect squares are

There are ten of them. Since there are slips in the bag, our fraction becomes $\frac{10}{100}$ .

If we cut a zero from both top and bottom, which is the same as dividing by ten, our final answer is

$\frac{1}{10}$ .

If an exam, there is only one right answer and four wrong answers, what's the probability of getting the one question right out of two?

$\frac{8}{25}$

$\frac{4}{25}$

$\frac{16}{25}$

$\frac{1}{25}$

$\frac{6}{25}$

Explanation

The probability of getting a question right is $\frac{1}{5}$ .

The probability of getting a question wrong is $\frac{4}{5}$ .

We are tempted to multiply both fractions to get an answer of $\frac{4}{25}$ . The issue is we don't know which question needs to be right.

So in this case, we multiply that fraction by to get an answer of $\frac{8}{25}$ .

I have six black balls, four green balls, eight blue balls, and two red balls. What is the probability I pick a blue ball first and then a black ball without replacement?

$\frac{12}{95}$

$\frac{7}{10}$

$\frac{3}{95}$

$\frac{14}{95}$

$\frac{7}{190}$

Explanation

Anytime we see the word and, we need to meet conditions. Therefore we need to multiply the probability.

The probability of getting a blue ball is $\frac{8}{20}$ or $\frac{2}{5}$ .

Next, the probability of getting a black ball is $\frac{6}{19}$ because it says without replacement, this means by drawing a black ball, we only have balls left.

Now, we multiply the fractions.

$\frac{2}{5}*\frac{6}{19}=\frac{12}{95}$

50 high school seniors were asked about their plans following graduation. 20 planned to attend an in state college. 12 planned to attend an out of state college. 8 did not plan to attend college and 10 were undecided.

What percent of this group of seniors is not planning to attend an in state college?

60%

40%

30%

20%

16%

Explanation

First, calculate the number of students not planning on attending an in state college.

Either subtract those planning on attending an in state college from the total...

50 - 20 = 30

...or add up all the other categories.

12 + 8 + 10 = 30

Then divide by the total number of students.

30/50 = 60%

In a standard deck of card, what is the probality of drawing either a face card or a red card?

$\frac{8}{13}$

$\frac{8}{52}$

$\frac{19}{52}$

$\frac{3}{26}$

$\frac{1}{2}$

Explanation

The probability of drawing a face card P(A) or a red card P(B) can be written as P(A or B), and be calculated using:

P(A or B) = P(A) + P(B) - P(A and B)

The first two terms on the right side of the equation are fairly straightforward. Since we'll be adding and subtracting fractions, I will keep the denominator the same and won't reduce the fractions until the end.

P(A) = 12/52

P(B) = 26/52 = 1/2

The third, P(A and B), can be calculated using:

P(A and B) = P(A) * P(B)

P(A and B) = 12/52 * 1/2 = 12/104 = 6/52

(Conversely, think there are 12 face cards in a deck of 52, half of which \[6\] are red.)

Back to our original equation!

P(A or B) = 12/52 + 26/52 - 6/52

= 32/52 = 8/13

Dylan rolls a six-sided die. What is the probability that he will roll a five on the first roll and a six on the second roll?

$\frac{1}{36}$

$\frac{1}{12}$

$\frac{1}{2}$

$\frac{1}{3}$

$\frac{1}{6}$

Explanation

To calculate the probability of two independent events occuring, you multiply the probabilities of each event occuring independently.

In this case, the probability of rolling any number is $\frac{1}{6}$ , so the probability of rolling this particular combination is:

$\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}$ .

Hector has a shelf of books. He wants to choose a random book to read. If there are 24 books on the shelf and he has an equal chance of choosing any particular one, what is the probability that he picks the third book from the left?

$\frac{1}{24}$

$\frac{3}{24}$

$\frac{1}{12}$

$\frac{5}{24}$

Explanation

If Hector has a shelf of books. He wants to choose a random book to read. If there are 24 books on the shelf and he has an equal chance of choosing any particular one, what is the probability that he picks the third book from the left?

This is a wordy problem, but if we cut through to what it's asking it is fairly simple.

We need to find the probability of an event.

The event is randomly choosing a book from a book shelf.

There are 24 books total, and we need to pick 1.

So, our probability will be

$P= \frac{#Desired}{Total#}=\frac{1}{24}$

So our answer is:

$\frac{1}{24}$

I have six black balls, four green balls, eight blue balls, and two red balls. What is the probability I pick either a blue or black ball?

$\frac{7}{10}$

$\frac{7}{20}$

$\frac{3}{5}$

$\frac{11}{20}$

$\frac{13}{20}$

Explanation

Anytime we see the word either, this means we are adding probabilities.

We are looking for a blue or black ball.

Our total number of blue and black balls is fourteen.

The total number of balls is twenty.

So our probability is:

$\frac{14}{20}=\frac{7}{10}$

If for each question on an exam, there is only one right answer and four wrong answers, what's the probability of getting the wrong answer for two questions?

$\frac{16}{25}$