PSAT Math : How to find the probability of an outcome

Study concepts, example questions & explanations for PSAT Math

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Example Questions

Example Question #81 : How To Find The Probability Of An Outcome

Joe has eight marbles: 3 Red, 2 Green, 2 Blue and 1 Yellow.  

If Joe puts the marbles into a bag and draws out four, what are the odds that he gets exactly one of each color? 

Possible Answers:

\frac{7}{1680}

\frac{12}{140}

\frac{1}{140}

\frac{1}{8}

\frac{3}{1024}

Correct answer:

\frac{1}{140}

Explanation:

The probabilities of each pick changes as fewer marbles remain in the bag.  Each selection order of the colors will yield the same probability, expressed as:

Example Question #82 : How To Find The Probability Of An Outcome

If you flip a fair coin four times in a row, what is the probability of getting at least one head?

Possible Answers:

\frac{1}{16}

\frac{1}{4}

\frac{3}{8}

\frac{15}{16}

\frac{5}{8}

Correct answer:

\frac{15}{16}

Explanation:

1-P(no\ heads)=P(at\ least\ one\ heads)

There are 16 different ways to flip a fair coin four times in a row.  There is only one way to get all tails.

P(no\ heads)=\frac{1}{16}

so the P(at\ least\ one\ heads)=1-\frac{1}{16}=\frac{15}{16}

Example Question #83 : How To Find The Probability Of An Outcome

When rolling two standard six-sided dice, what is the probability of getting five or less?

Possible Answers:

\frac{1}{6}

\frac{3}{8}

\frac{5}{16}

\frac{7}{12}

\frac{5}{18}

Correct answer:

\frac{5}{18}

Explanation:

The sample space for rolling two six-sided dice is 36.

Counting the wanted outcomes gives:

2:  1,1 (1)

3: 1,2 and 2,1 (2)

4: 1,3 and 2,2 and 3,1 (3)

5: 1,4 and 2,3 and 3,2 and 4,1 (4)

So there are 10 ways to get a five or less.

Thus the probability of getting a five or less is \frac{10}{36}=\frac{5}{18}.

Example Question #84 : How To Find The Probability Of An Outcome

How many different ways can five books be lined up on a shelf?

Possible Answers:

225

120

25

90

75

Correct answer:

120

Explanation:

Order matters, so we use permutations:  _{5}^{5}\textrm{P}5\times 4\times 3\times 2\times 1=120.

Example Question #85 : How To Find The Probability Of An Outcome

You have a box of 90 colored scrunchies. Half of the scrunchies are black, one third of the scrunchies are white, one ninth of the scrunchies are blue, and the rest are green. You pull the scrunchies from the box at random.

The first scrunchie you pick up is blue. The second scrunchie is green. What is the probability that the third scrunchie you pick up will be black?

Possible Answers:

\frac{43}{88}

\frac{1}{2}

\frac{45}{88}

None of the other answers

\frac{43}{90}

Correct answer:

\frac{45}{88}

Explanation:

Half of the scrunchies are black, so

90\times \frac{1}{2}=45\hspace{1 mm}black\hspace{1 mm}scrunchies

One third of the scrunchies are white

90\times \frac{1}{3}=30\hspace{1 mm}white\hspace{1 mm}scrunchies

One ninth of the scrunchies are blue

90\times \frac{1}{9}=10\hspace{1 mm}blue\hspace{1 mm}scrunchies

And the rest are green:

90-45-30-10=5\hspace{1 mm}green\hspace{1 mm}scrunchies

If we have already drawn two, our total amount of scrunchies is now 88, so the probability of pulling a black scrunchie will be:

\frac{45}{88}

 

 

Example Question #171 : Statistics

There is a classroom of 60 students. \frac{7}{12} of the students are wearing red shirts, \frac{1}{3} of the students are wearing pink shirts, and the rest of the students are wearing orange shirts. What is the probability of randomly selecting the name of a student wearing an orange shirt?

Possible Answers:

8.3\%

\frac{1}{12}

\frac{2}{25}

\frac{1}{6}

None of the available answers

Correct answer:

\frac{1}{12}

Explanation:

\frac{7}{12} of the students are wearing red shirts, and

\frac{7}{12}\times 60=\frac{7}{1}\times 5=35\hspace{1 mm}students

One third of the students are wearing pink shirts

\frac{1}{3}\times 60=\frac{1}{1}\times 20=20\hspace{1 mm}students

The remaining students are wearing orange

60-35-20=5\hspace{1 mm}students

The probability of randomly selecting a student wearing an orange shirt is

\frac{5}{60}=\frac{1}{12}

Example Question #86 : How To Find The Probability Of An Outcome

A bag of colored candies has ten red candies, 14 purple candies, 12 orange candies, and 4 yellow candies. What is the ratio of yellow to orange candies in the bag?

Possible Answers:

Correct answer:

Explanation:

There are four yellow candies, and 12 orange candies, so the raio is:

The ratio of yellow to orange candies is 1:3

Example Question #87 : How To Find The Probability Of An Outcome

 of the population in Town A is NOT Asian. In addition,  of the Asian population in Town A is male. A random person from Town A is selected. What is the probability that the person selected is both Asian and female.

Possible Answers:

Correct answer:

Explanation:

Recall:

Consequently, the probability that the person selected is Asian is:

Similarly, the probability that a randomly selected Asian person is also female is:

Finally, the probability of 2 unrelated events occuring is equal to the product of the individual probabilities of the 2 events. Therefore the probability of selecting an Asian female is

Example Question #88 : How To Find The Probability Of An Outcome

Fred wants to put together outfits that he can wear for the day.  He can wear any type of outfit for school, but he needs to wear formal attire for dinner later.  He can either choose to wear something all day or choose to change clothes before going out to dinner.

If Fred has 2 formal shirts and 3 informal shirts, 1 pair of formal pants and 2 pairs of informal pants, and 1 pair of formal shoes and 1 pair of informal shoes, how many different ways could Fred appear for school and then dinner, allowing for changing in between?

Possible Answers:

Correct answer:

Explanation:

Fred can wear anything to school. This means he can wear either formal things or informal things.  He has 3 informal shirts and 2 formal shirts.  This means he has a total of 5 shirts to choose from when deciding what to wear.  Similarly, he has 3 pairs of pants and 2 pairs of shoes to choose from.  After choosing a shirt (5 options), he chooses pants (3 options) and shoes (2 options) resulting in  choices for outfits to wear to school.

Fred must wear formal attire to the dinner.  This means that Fred can only choose between his 2 formal shirts, his formal pair of pants, and his formal shoes.  Since he only has a choice of shirts, he ends up only having 2 choices to make in outfits.

To figure out how many ways he can dress throughout the day, we simply choose an outfit for school and an outfit for dinner.  We have 30 choices for what to wear to school and 2 choices to wear to dinner, giving us choices in total.

Example Question #89 : How To Find The Probability Of An Outcome

A school takes a poll of their students and 85% respond.  They find that, of those who responded, 45% are male and 55% are female.  Of the males who responded to the poll, 80% said they would prefer more money invested in extracurriculars rather than the core educational classes.  100% of female respondents agreed that more money should be invested in extracurriculars.

If the school has 2000 students, which of the following could possibly be true?

Possible Answers:

A greater number of male students prefer core classes to extracurriculars.

Less than  of the school believe that the school should increase spending for extracurriculars.

 More males than females believe there should be increased spending on extracurriculars.

The majority of students believe the school should increase funding of core classes.

The number of males who think that the school should have more extracurriculars is greater than 900.

Correct answer:

The number of males who think that the school should have more extracurriculars is greater than 900.

Explanation:

Firstly, when taking a percentage, we can simply multiply by the proper decimal, where multiplying by 1 is 100%.

The school has 2,000 students.  Since only 85% responded, we have  students that were not accounted for. 

Of the 1700 who responded,  are male.  Of these students,  believe that there should be more invested in extracurriculars.

 

The key is recognizing that we can 'control' the 300 votes that are unaccounted for.  These students were not counted, but might be male or female, and might vote for or against increased spending on extracurriculars.

Thus while we initially have 612 males believing in extracurriculars, it might be true that the remaining 300 uncounted votes were all males that believed in increasing spending on extracurriculars, resulting in up to 912 males voting for it.

 

All the other options are entirely impossible. 

We know that  are in favor of spending on extracurriculars.  This means that the majority of students believe in increased spending on extracurriculars not on core classes, and that more than  of the school believe that this spending should be increased.

We also know that, even should we have the maximum possible of 912 males in favor of increasing spending on extracurriculars, the 935 females who are in favor are still greater.

Finally, the number of male students who prefer extracurriculars is currently 612.  Even should all 300 remaining voters be males in favor of core classes, we end up with  in favor of core classes vs 612 in favor of extracurriculars.

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