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PSAT Math : How to find the probability of an outcome

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Example Questions

Example Question #51 : Calculating Discrete Probability

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

Possible Answers:

$\frac{25}{52}$

$\frac{9}{26}$

$\frac{11}{26}$

$\frac{3}{52}$

$\frac{156}{2704}$

Correct answer:

$\frac{11}{26}$

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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Example Question #44 : Probability

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

Possible Answers:

$\frac{1}{4}$

$\frac{1}{16}$

$\frac{11}{16}$

$\frac{5}{16}$

$\frac{3}{4}$

Correct answer:

$\frac{5}{16}$

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:

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

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Example Question #41 : Probability

Rolling a four-sided dice, what is the probability of rolling a three times out of four?

Possible Answers:

$\frac{3}{64}$

$\frac{13}{256}$

$\frac{3}{256}$

$\frac{243}{256}$

$\frac{3}{128}$

Correct answer:

$\frac{3}{64}$

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 rolling a four), and is the probability of success (one in four).

$P(4,3)=\frac{4!}{(4-3)!3!}(\frac{1}{4})^3(1-\frac{1}{4})^{4-3}=4(\frac{1}{64})(\frac{3}{4})=\frac{3}{64}$

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Example Question #43 : How To Find The Probability Of An Outcome

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

Possible Answers:

$\frac{1}{7}$

$\frac{1}{128}$

$\frac{127}{128}$

$\frac{6}{7}$

$\frac{255}{256}$

Correct answer:

$\frac{127}{128}$

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:

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

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Example Question #1548 : Psat Mathematics

Set A: (A,C,T,Q,U,X,Z)

Set B: (I,A,W,X,E,R)

One letter is picked from Set A and Set B. What is the probability of picking two consonants?

Possible Answers:

$\frac{4}{7}$

$\frac{5}{7}$

$\frac{3}{14}$

$\frac{5}{14}$

$\frac{2}{14}$

Correct answer:

$\frac{5}{14}$

Explanation:

Set A: (A,C,T,Q,U,X,Z)

Set B: (I,A,W,X,E,R)

In Set A, there are five consonants out of a total of seven letters, so the probability of drawing a consonant from Set A is $\frac{5}{7}$ .

In Set B, there are three consonants out of a total of six letters, so the probability of drawing a consonant from Set B is $\frac{1}{2}$ .

The question asks for the probability of drawing two consonants, meaning the probability of drawing a constant from Set A and Set B, so probability of the intersection of the two events is the product of the two probabilities:

$\frac{5}{7}(\frac{1}{2})=\frac{5}{14}$

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Example Question #52 : Calculating Discrete Probability

Set A: (A,C,T,Q,U,X,Z)

Set B: (I,A,W,X,E,R)

One letter is picked from Set A and Set B. What is the probability of picking at least one consonant?

Possible Answers:

$\frac{5}{13}$

$\frac{1}{7}$

$\frac{5}{14}$

$\frac{6}{7}$

$\frac{17}{14}$

Correct answer:

$\frac{6}{7}$

Explanation:

Set A: (A,C,T,Q,U,X,Z)

Set B: (I,A,W,X,E,R)

In Set A, there are five consonants out of a total of seven letters, so the probability of drawing a consonant from Set A is $\frac{5}{7}$ .

In Set B, there are three consonants out of a total of six letters, so the probability of drawing a consonant from Set B is $\frac{1}{2}$ .

The question asks for the probability of drawing at least one consonant, which can be interpreted as a union of events. To calculate the probability of a union, sum the probability of each event and subtract the intersection:

$P(A\bigcup B)=P(A)+P(B)-P(A\bigcap B)$

The interesection is:

$P(A\bigcap B)=P(A)P(B)=\frac{5}{7}(\frac{1}{2})=\frac{5}{14}$

So, we can find the probability of drawing at least one consonant:

$P(A\bigcup B)= \frac{5}{7}+\frac{1}{2}-\frac{5}{14}=\frac{10}{14}+\frac{7}{14}-\frac{5}{14}=\frac{6}{7}$

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Example Question #201 : Statistics

Set A: (A,C,T,Q,U,X,Z)

Set B: (I,A,W,X,E,R)

One letter is drawn from Set A, and one from Set B. What is the probability of drawing a matching pair of letters?

Possible Answers:

$\frac{2}{13}$

$\frac{4}{13}$

$\frac{1}{7}$

$\frac{1}{21}$

$\frac{1}{42}$

Correct answer:

$\frac{1}{21}$

Explanation:

Set A: (A,C,T,Q,U,X,Z)

Set B: (I,A,W,X,E,R)

Between Set A and Set B, there are two potential matching pairs of letters: AA and XX. The amount of possible combinations is the number of values in Set A, multiplied by the number of values in Set B, 7(6)=42 .

Therefore, the probability of drawing a matching set is:

$\frac{2}{42}=\frac{1}{21}$

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Example Question #164 : Data Analysis

In a particular high school, 200 students are freshmen, 150 students are sophomores, 250 students are juniors, and 100 students are seniors. Twenty percent of freshmen are in honors classes, ten percent of sophomores are in honors classes, twelve percent of juniors are in honors classes, and thirty percent of seniors are in honors classes.

If a student is chosen at random, what is the probability that that student will be a student who attends honors classes?

Possible Answers:

$\frac{5}{7}$

$\frac{9}{50}$

$\frac{9}{100}$

$\frac{23}{70}$

$\frac{23}{140}$

Correct answer:

$\frac{23}{140}$

Explanation:

First calculate the number of students:

200+150+250+100=700

The probability of drawing an honors student will then be the total number of honors students divided by the total number of students attending the school:

$\frac{200(.20)+(150)(.10)+(250)(.12)+(100)(0.30)}{700}$

$\frac{115}{700}=\frac{23}{140}$

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Example Question #61 : Discrete Probability

If a student is chosen at random, what is the probability that that student will be a senior student and a student who does not attend honors classes?

Possible Answers:

$\frac{3}{10}$

$\frac{4}{35}$

$\frac{2}{5}$

$\frac{3}{70}$

$\frac{1}{10}$

Correct answer:

$\frac{1}{10}$

Explanation:

First calculate the number of students:

200+150+250+100=700

The percentage of seniors that do not attend honors classes is:

(1-.30)=.70

Therefore, the probability of selecting a student who is a senior and one who does not attend honors classes is:

$\frac{100}{700}(.70)=\frac{1}{10}$

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Example Question #221 : Data Analysis

11 cards are placed into a box numbered 5 - 15. If one card is randomly drawn from the box, what is the probabiltiy that a prime number will be on the card?

Possible Answers:

2/5

6/11

1/2

3/5

4/11

Correct answer:

4/11

Explanation:

Possible numbers 5,6,7,8,9,10,11,12,13,14,15 (11 total numbers)

Prime numbers are 5,7,11,13 (4 total prime numbers)

totalnumber of prime numbers/ total numbers in box

Answer = 4/11

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PSAT Math : How to find the probability of an outcome

Study concepts, example questions & explanations for PSAT Math

All PSAT Math Resources

Example Questions

Example Question #51 : Calculating Discrete Probability

Example Question #44 : Probability

Example Question #41 : Probability

Example Question #43 : How To Find The Probability Of An Outcome

Example Question #1548 : Psat Mathematics

Example Question #52 : Calculating Discrete Probability

Example Question #201 : Statistics

Example Question #164 : Data Analysis

Example Question #61 : Discrete Probability

Example Question #221 : Data Analysis

All PSAT Math Resources

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