Conditional Probability Practice Questions — GMAT

Question 1Question →

An analyst at an investment firm evaluates a portfolio of $160$ corporate bonds for two potential risk factors: credit rating downgrade risk and liquidity risk. The evaluation reveals that $60$ bonds have credit rating downgrade risk, $72$ bonds have liquidity risk, and $64$ bonds have neither risk factor. If a bond is selected at random from those in the portfolio that have at least one of the two risk factors, what is the probability that it has credit rating downgrade risk? Express your answer as a decimal.

Show answer & explanation

Answer: 0.625

Answer

The probability is 0.625 (or 5/8).

To calculate the probability that a bond has credit rating downgrade risk given that it has at least one risk factor, the sample space must be restricted to bonds with at least one risk factor. Out of

160

bonds,

64

have neither risk factor, leaving

160 - 64 = 96

bonds with at least one risk factor. All

60

bonds with credit rating downgrade risk are part of this group. The required conditional probability is

\frac{60}{96} = \frac{5}{8} = 0.625

.

Step-by-Step Solution

1

Calculate the size of the restricted sample space (bonds with at least one risk factor).

Total bonds with at least one risk factor = 160 - 64 = 96 bonds.

The condition specifies that the selection is made only from bonds having at least one risk factor.

2

Identify the number of favorable outcomes within this restricted sample space.

Number of bonds with credit rating downgrade risk = 60.

All 60 bonds with credit rating downgrade risk inherently possess at least one risk factor, so they lie entirely within the restricted sample space.

3

Compute the conditional probability P(Downgrade Risk | At Least One Risk).

60 / 96 = 5 / 8 = 0.625.

Conditional probability requires dividing the count of favorable outcomes by the count of the restricted sample space.

Key Concept

Conditional Probability and Sample Space Restriction

Question 2Question →

An integer is selected at random from the set of all positive integers less than or equal to 40. Given that the selected integer is a multiple of 3, what is the probability that it is also a multiple of 4?

\frac{3}{40}

\frac{3}{13}

\frac{3}{10}

\frac{1}{4}

\frac{4}{13}

Show answer & explanation

Answer: $\frac{3}{13}$

Answer

The probability that the selected integer is a multiple of 4, given that it is a multiple of 3, is

\frac{3}{13}

.

The condition 'given that the selected integer is a multiple of 3' restricts the sample space to the 13 positive integers up to 40 that are divisible by 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, and 39. Among these 13 numbers, those that are also multiples of 4 are multiples of 12, namely 12, 24, and 36 (3 numbers). Therefore, the conditional probability is

\frac{3}{13}

.

Step-by-Step Solution

1

Identify the restricted sample space defined by the condition.

The positive integers less than or equal to 40 that are multiples of 3 are: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, and 39. There are 13 such integers.

For conditional probability

P(A \mid B)

, the sample space must be restricted to all outcomes satisfying condition

B

(being a multiple of 3).

2

Identify the favorable outcomes within the restricted sample space.

Integers that are multiples of both 3 and 4 must be multiples of 12. Among the 13 multiples of 3, those that are also multiples of 12 are: 12, 24, and 36. There are 3 such integers.

The numerator of

P(A \mid B)

counts the elements in the intersection

A \cap B

.

3

Calculate the conditional probability.

P(\text{Multiple of 4} \mid \text{Multiple of 3}) = \frac{3}{13}

.

Divide the number of favorable outcomes (3) by the total number of outcomes in the restricted sample space (13).

Key Concept

Conditional Probability with Restricted Sample Space

Estimated Time:1m 30s

Question 3Question →

A laboratory tested $200$ synthetic compound samples for two properties: thermal stability and chemical resistance. Among the samples, $120$ exhibited thermal stability, $90$ exhibited chemical resistance, and $50$ exhibited neither property. If a sample is selected at random from those that exhibited thermal stability, what is the probability that it also exhibited chemical resistance?

Show answer & explanation

Answer: 0.5

Answer

0.5

The conditional probability of selecting a sample with chemical resistance given that it has thermal stability is found by dividing the number of samples with both properties (

60

) by the total number of samples with thermal stability (

120

), giving

\frac{60}{120} = 0.5

.

Step-by-Step Solution

1

Find the total number of samples exhibiting at least one of the two properties.

Since

50

out of

200

samples exhibited neither property, the number of samples exhibiting at least one property is

200 - 50 = 150

.

The total population consists of samples exhibiting at least one property plus samples exhibiting neither property.

2

Calculate the number of samples exhibiting both thermal stability (

T

) and chemical resistance (

C

).

Using the inclusion-exclusion principle

|T \cup C| = |T| + |C| - |T \cap C|

, we have

150 = 120 + 90 - |T \cap C|

, which yields

|T \cap C| = 60

.

Overlapping sets require subtracting the intersection to avoid double-counting elements.

3

Compute the conditional probability

P(C|T)

.

P(C|T) = \frac{|T \cap C|}{|T|} = \frac{60}{120} = 0.5

.

The given condition restricts the sample space to only the

120

samples exhibiting thermal stability.

Key Concept

Conditional probability restricts the sample space to the given condition's outcome space:

P(A|B) = \frac{|A \cap B|}{|B|}

.

Estimated Time:1m 30s

Question 4Question →

In a group of $100$ students, $60$ students study Spanish and $40$ students study French. Among the $60$ students studying Spanish, $15$ also study French. If a student who studies Spanish is selected at random, what is the probability that the selected student also studies French?

Show answer & explanation

Answer: 0.25

Answer

The probability that a randomly selected Spanish-studying student also studies French is 0.25.

Since the student is chosen from the group of

60

Spanish-studying students, the sample space is restricted to

60

. Within this subset,

15

students study French. The conditional probability is therefore

\frac{15}{60} = 0.25

.

Step-by-Step Solution

1

Identify the total number of outcomes in the restricted sample space.

The sample space is restricted to students studying Spanish:

n(\text{Spanish}) = 60

.

The question specifies that the student is chosen from those who study Spanish.

2

Identify the number of favorable outcomes within this restricted sample space.

The number of students studying both Spanish and French is

n(\text{Spanish} \cap \text{French}) = 15

.

We need the count of students who satisfy both the given condition and the target event.

3

Calculate the conditional probability.

P(\text{French} | \text{Spanish}) = \frac{15}{60} = 0.25

.

Conditional probability is calculated by dividing the intersection count by the given condition's total count.

Key Concept

Conditional Probability:

P(A|B) = \frac{n(A \cap B)}{n(B)}

Question 5Question →

A box contains 12 cards numbered consecutively from 1 through 12. Two cards are selected at random from the box without replacement. Given that the sum of the numbers on the two selected cards is even, what is the probability that at least one of the selected cards has a prime number on it?

\frac{19}{30}

\frac{19}{66}

\frac{2}{3}

\frac{7}{15}

\frac{15}{22}

Show answer & explanation

Answer: $\frac{19}{30}$

Answer

The conditional probability that at least one selected card has a prime number on it, given that their sum is even, is

\frac{19}{30}

.

The option offering 19/30 is correct. Given that the sum of the two cards is even, both cards must be even or both cards must be odd. Choosing 2 even cards from 6 available gives 15 pairs, and choosing 2 odd cards from 6 available gives 15 pairs, making 30 possible pairs in total for the restricted sample space. Among the even cards, 2 is prime while 4, 6, 8, 10, and 12 are non-prime (5 numbers). Among the odd cards, 3, 5, 7, and 11 are prime (4 numbers) while 1 and 9 are non-prime (2 numbers). The pairs containing no primes consist of 2 non-prime evens (10 pairs) and 2 non-prime odds (1 pair), giving 11 non-prime pairs. Subtracting from 30 yields 19 pairs with at least one prime. Thus, the conditional probability is 19/30.

Step-by-Step Solution

1

Determine the restricted sample space (Condition B: Sum of two cards is even)

Total outcomes in Condition B = 30

The sum of two integers is even if both are even or both are odd. Among numbers 1 to 12, there are 6 even numbers ({2, 4, 6, 8, 10, 12}) and 6 odd numbers ({1, 3, 5, 7, 9, 11}). The number of ways to pick 2 even cards is

\binom{6}{2} = 15

, and 2 odd cards is

\binom{6}{2} = 15

. Total pairs with an even sum =

15 + 15 = 30

.

2

Categorize the numbers 1 through 12 by parity and primality

Prime evens = {2} (1 number); Non-prime evens = {4, 6, 8, 10, 12} (5 numbers); Prime odds = {3, 5, 7, 11} (4 numbers); Non-prime odds = {1, 9} (2 numbers)

Note that 1 is not a prime number, and 2 is the only even prime number.

3

Count the number of pairs in the restricted sample space with NO prime numbers

11 non-prime pairs

Pairs of two evens with no primes come from non-prime evens:

\binom{5}{2} = 10

pairs. Pairs of two odds with no primes come from non-prime odds:

\binom{2}{2} = 1

pair. Total non-prime pairs =

10 + 1 = 11

.

4

Calculate favorable outcomes (Event A ∩ B) and the conditional probability

P(A|B) = 19/30

Favorable pairs with at least one prime =

30 - 11 = 19

. Therefore,

P(\text{At least one prime} \mid \text{Even sum}) = \frac{19}{30}

.

Key Concept

Conditional Probability with Restricted Sample Space

Estimated Time:2m 0s

Question 6Question →

A software company audited $120$ applications for compliance with two protocols: Accessibility (Protocol A) and Security (Protocol B). The audit revealed that $75$ applications complied with Protocol A, $60$ applications complied with Protocol B, and $35$ applications complied with neither protocol. If an application selected at random from the audited group is known to comply with Protocol A, what is the probability that it also complies with Protocol B?

\frac{5}{12}

\frac{2}{3}

\frac{5}{6}

\frac{2}{15}

\frac{17}{24}

Show answer & explanation

Answer: $\frac{2}{3}$

Answer

\frac{2}{3}

Out of the 120 total applications, 35 comply with neither protocol, leaving 85 that comply with at least one. Using the inclusion-exclusion principle (

75 + 60 - 85 = 50

), exactly 50 applications comply with both protocols. Since the application is already known to comply with Protocol A, the denominator is restricted to the 75 applications in Protocol A. The conditional probability is therefore

50 / 75 = 2/3

.

Step-by-Step Solution

1

Determine the number of applications that comply with at least one protocol.

Total applications minus those complying with neither:

120 - 35 = 85

.

The total sample space is partitioned into applications complying with at least one protocol and those complying with neither.

2

Find the number of applications complying with both Protocol A and Protocol B using the Principle of Inclusion-Exclusion.

|A \cap B| = |A| + |B| - |A \cup B| = 75 + 60 - 85 = 50

.

Summing the counts of Protocol A and Protocol B counts applications in both protocols twice.

3

Calculate the conditional probability

P(B|A)

.

P(B|A) = \frac{|A \cap B|}{|A|} = \frac{50}{75} = \frac{2}{3}

.

Given that the selected application complies with Protocol A, the sample space is restricted to

|A| = 75

.

Key Concept

Conditional Probability with Overlapping Sets

Estimated Time:1m 30s

Conditional Probability

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Common Mistakes

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Common Mistakes

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Common Mistakes

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Common Mistakes

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