GRE Subject Test: Math : Probability & Statistics

Study concepts, example questions & explanations for GRE Subject Test: Math

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Example Questions

Example Question #23 : Permutation / Combination

If there are \(\displaystyle 20\) students in a class and \(\displaystyle 2\) people are randomly choosen to become class representatives, how many different ways can the representatives be chosen?

Possible Answers:

\(\displaystyle 180\)

\(\displaystyle 190\)

\(\displaystyle 200\)

\(\displaystyle 210\)

\(\displaystyle 222\)

Correct answer:

\(\displaystyle 190\)

Explanation:

To solve this problem, we must understand the concept of combination/permutations. A combination is used when the order doesn't matter while a permutation is used when order matters. In this problem, the two class representatives are randomly chosen, therefore it doesn't matter what order the representative is chosen in, the end result is the same.  The general formula for combinations is \(\displaystyle C(n,k)=\frac{n!}{(n-k)!k!}\), where \(\displaystyle n\) is the number of things you have and \(\displaystyle k\) is the things you want to combine.

 

Plugging in choosing 2 people from a group of 20, we find

\(\displaystyle C(20,2)=\frac{20!}{(20-2)!2!}=190\).  Therefore there are a \(\displaystyle 190\) different ways to choose the \(\displaystyle 2\) class representatives.

Example Question #1 : Combinations

There are eight possible flavors of curry at a particular restaurant.

Quantity A: Number of possible combinations if four unique curries are selected.

Quantity B: Number of possible combinations if five unique curries are selected.

Possible Answers:

The two quantities are equal

The relationship cannot be determined.

Quantity B is greater.

Quantity A is greater.

Correct answer:

Quantity A is greater.

Explanation:

The number of potential combinations for \(\displaystyle k\) selections made from \(\displaystyle n\) possible options is

\(\displaystyle C(n,k)=\frac{n!}{(n-k)!k!}\)

Quantity A:

\(\displaystyle C(8,4)=\frac{8!}{(8-4)!4!}=70\)

Quantity B:

\(\displaystyle C(8,5)=\frac{8!}{(8-5)!5!}=56\)

Quantity A is greater.

Example Question #2 : Combinations

Quantity A: The number of possible combinations if four unique choices are made from ten possible options.

Quantity B: The number of possible permutations if two unique choices are made from ten possible options.

Possible Answers:

The two quantities are equal.

Quantity B is greater.

The relationship cannot be established.

Quantity A is greater.

Correct answer:

Quantity A is greater.

Explanation:

For \(\displaystyle k\) choices made from \(\displaystyle n\) possible options, the number of potential combinations (order does not matter) is

\(\displaystyle C(n,k)=\frac{n!}{(n-k)!k!}\)

And the number of potential permutations (order matters) is

\(\displaystyle P(n,k)=\frac{n!}{(n-k)!}\)

Quantity A:

\(\displaystyle C(10,4)=\frac{10!}{(10-4)!4!}=210\)

Quantity B:

\(\displaystyle P(10,2)=\frac{10!}{(10-2)!}=90\)

Quantity A is greater.

Example Question #5 : Combinations

There are \(\displaystyle 10\) possible flavor options at an ice cream shop.

\(\displaystyle \textup{Quantity A: The number of possible combinations if three unique flavors are selected.}\)

\(\displaystyle \textup{Quantity B: The number of possible combinations if five unique flavors are selected.}\)

Possible Answers:

\(\displaystyle \textup{One of the numbers is not a real number}\)

\(\displaystyle \textup{The relationship cannot be determined}\)

\(\displaystyle \textup{Quantity B is greater}\)

\(\displaystyle \textup{Quantity A is greater}\)

\(\displaystyle \textup{The two quantities are equal}\)

Correct answer:

\(\displaystyle \textup{Quantity B is greater}\)

Explanation:

When dealing with combinations, the number of possible combinations when selecting \(\displaystyle k\) choices out of \(\displaystyle n\) options is:

\(\displaystyle C(n,k)=\frac{n!}{(n-k)! k!}\)

For Quantity A, the number of combinations is:

\(\displaystyle C(10,3)=\frac{10!}{(10-3)!3!}\)

\(\displaystyle C(10,3)=120\)

For Quantity B, the number of combinations is:

\(\displaystyle C(10,5)=\frac{10!}{(10-5)!5!}\)

\(\displaystyle C(10,5)=252\)

Quantity B is greater.

Example Question #3 : Combinations

Quantity A: The number of potential combinations given two choices made from ten options.

Quantity B: The number of potential combinations given four choices made from twenty options.

Possible Answers:

The two quantities are equal.

The relationship cannot be determined.

Quantity A is larger.

Quantity B is larger.

Correct answer:

Quantity B is larger.

Explanation:

Since in this problem we're dealing with combinations, the order of selection does not matter.

With \(\displaystyle k\) selections made from \(\displaystyle n!\) potential options, the total number of possible combinations is

\(\displaystyle C(n,k)=\frac{n!}{(n-k)!k!}\)

Quantity A:

\(\displaystyle C(10,2)=\frac{10!}{(10-2)!2!}=45\)

Quantity B:

\(\displaystyle C(20,4)=\frac{20!}{(20-4)!4!}=4845\)

Quantity B is larger.

Example Question #4 : Combinations

Quantity A: The number of combinations if five choices are made from ten options.

Quantity B: The number of combinations if two choices are made from twenty options.

Possible Answers:

The two quantities are equal.

Quantity B is greater.

Quantity A is greater.

The relationship cannot be determined.

Correct answer:

Quantity A is greater.

Explanation:

Since we're dealing with combinations in this problem, the order of selection does not matter.

With \(\displaystyle k\) selections made from \(\displaystyle n!\) potential options, the total number of possible combinations is

\(\displaystyle C(n,k)=\frac{n!}{(n-k)!k!}\)

Quantity A:

\(\displaystyle C(10,5)=\frac{10!}{(10-5)!5!}=252\)

Quantity B:

\(\displaystyle C(20,2)=\frac{20!}{(20-2)!2!}=190\)

Quantity A is greater.

Example Question #5 : Combinations

Rachel is buying ice cream for a sundae. If there are twelve ice cream choices, how many scoops will give the maximum possible number of unique sundaes?

Possible Answers:

\(\displaystyle 7\)

\(\displaystyle 9\)

\(\displaystyle 8\)

\(\displaystyle 6\)

\(\displaystyle 10\)

Correct answer:

\(\displaystyle 6\)

Explanation:

Since in this problem the order of selection does not matter, we're dealing with combinations.

With \(\displaystyle k\) selections made from \(\displaystyle n!\) potential options, the total number of possible combinations is

\(\displaystyle C(n,k)=\frac{n!}{(n-k)!k!}\)

In terms of finding the maximum number of combinations, the value of \(\displaystyle k\) should be 

\(\displaystyle k_{max}=\frac{n_{even}}{2};k_{max}=\frac{n_{odd}\pm 1}{2}\)

Since there are twelve options, a selection of six scoops will give the maximum number of combinations.

\(\displaystyle C(12,6)=\frac{12!}{(12-6)!6!}=924\)

\(\displaystyle C(12,7)=792\)

\(\displaystyle C(12,8)=495\)

\(\displaystyle C(12,9)=220\)

\(\displaystyle C(12,10)=66\)

Example Question #9 : Combinations

A coach must choose \(\displaystyle 5\) starters from a team of \(\displaystyle 11\) players. How many ways can the coach choose the starters?

Possible Answers:

\(\displaystyle 484\)

\(\displaystyle 462\)

\(\displaystyle 324\)

\(\displaystyle 492\)

Correct answer:

\(\displaystyle 462\)

Explanation:

Step 1: We need to read the question carefully. Order does not matter here.

Step 2: Order does not matter, so we need to use Combination.

Step 3: The combination formula is \(\displaystyle nCr=\frac{n!}{r!(n-r)!}\).

Step 4: We need to find the value of \(\displaystyle n\) and \(\displaystyle r\).

The value of \(\displaystyle n\) is how many players the coach can choose from, so \(\displaystyle n=11\)
The value of \(\displaystyle r\) is how many players that the coach can choose at one time, so \(\displaystyle r=5\).

Step 5: Plug in the values of n and r into the equation in step 2:

\(\displaystyle nCr=\frac{11!}{5!(11-5)!}\)

Step 6. Simplify the equation in step 5. The "!" means that I multiply that number by every other number below until 1.

\(\displaystyle nCr=\frac{11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{5\cdot 4\cdot 3\cdot 2\cdot 1\cdot 6 \cdot5\cdot 4\cdot 3\cdot 2\cdot 1}\)

Step 7: Cross out any terms that are on both the top and the bottom. We see \(\displaystyle 1-6\) is on top and bottom.

\(\displaystyle nCr=\frac{11\cdot 10\cdot 9\cdot 8\cdot 7}{5\cdot4\cdot3\cdot2\cdot1}\)

Step 8: Cross out \(\displaystyle 5\cdot2\) in the denominator with \(\displaystyle 10\) in the numerator. Rewrite.

\(\displaystyle nCr=\frac{11\cdot 9\cdot 8\cdot 7}{4\cdot 3\cdot 1}\)

Step 9: Divide \(\displaystyle 8\) in the numerator by \(\displaystyle 4\) in the denominator.

\(\displaystyle nCr=\frac{11\cdot 9\cdot 2\cdot 7}{3\cdot 1}\)

Step 10: Divide \(\displaystyle 9\) in the numerator by \(\displaystyle 3\) in the denominator.

\(\displaystyle nCr=11\cdot 3\cdot2\cdot7\)

Step 11: Multiply the right side

\(\displaystyle 11C5=462\)

There are 462 ways that the coach can choose 5 players out of 11 players on the bench.

Example Question #6 : Combinations

How many ways can a coach choose \(\displaystyle 6\) players to play on the field out of a bench of \(\displaystyle 13\) players?

Possible Answers:

\(\displaystyle 13!-6!\)

\(\displaystyle 13C6\)

\(\displaystyle 13P6\)

\(\displaystyle \frac {13!}{7!-6!}\)

Correct answer:

\(\displaystyle 13C6\)

Explanation:

Step 1: Read the question carefully. Look for hints of restrictions..

There are no order in which players can be chosen, which goes against the definition of Permutation. Permutation is the arrangement of objects by way of order.. If it's not permutation, it's Combination.

Step 2: Write what we know down..

Total Players \(\displaystyle (n)\)=\(\displaystyle 13\)
Choosing # of players \(\displaystyle (r)\)=\(\displaystyle 6\)..

Step 3: Plug in the numbers to the formula: \(\displaystyle nCr\)..

We ger 13C6. 

There is no need to evaluate this expression...

Example Question #11 : Combinations

Find \(\displaystyle C(8,5)\).

Possible Answers:

\(\displaystyle 54\)

\(\displaystyle 52\)

\(\displaystyle 56\)

\(\displaystyle 48\)

Correct answer:

\(\displaystyle 56\)

Explanation:

There are two types of statistical calculations that are used when dealing with ordering a number of objects. When the order does not matter it is known as a combination and denoted by a C.

\(\displaystyle C(n,k)=\frac{n!}{k!(n-k)!}\)

Thus the formula for this particular combination is,

\(\displaystyle C (8,5) = P \frac{(8,5)}{5!}\)

\(\displaystyle \frac{8\times 7\times 6\times 5\times 4}{5\times 4\times 3\times 2\times 1} =\)

The \(\displaystyle 5 \times 4\) will cancel out because it is in the numerator and denominator,

\(\displaystyle \frac{8\times 7\times 6}{3\times 2\times 1} = \frac{336}{6} = 56\).

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