Practice: Probability Trees and Conditional Expectations

Quantitative Methods. 13 question(s) in this unit's pool (0 above the exam). Free up to ten a day; the coach picks which ones based on what you have already answered and when each is next due.

Quantitative MethodsProbability Trees and Conditional Expectations
Your state on this unit Not started

Back to your map

Today's practice

Pick an answer, say how sure you are, then reveal. Every wrong choice gets its own explanation. Questions you have already answered correctly and confidently stay out of the way until they are due for review again.

Question 1Exam level

An analyst estimates there is a 40% probability that a company will meet earnings expectations and a 30% probability that the company will announce a new product line. The probability that the company both meets earnings AND announces a new product is 15%. The probability that the company meets earnings OR announces a new product is closest to:

How sure are you?

Correct: B. P(A or B) = P(A) + P(B) - P(A and B) = 0.40 + 0.30 - 0.15 = 0.55.
A. You might add 0.40 + 0.30 = 0.70, forgetting the overlap subtraction entirely. Events are not mutually exclusive (P(A and B) = 0.15 ≠ 0), so the overlap must be subtracted.
C. 0.45 does not satisfy the addition rule: P(A or B) = P(A) + P(B) - P(A and B) = 0.40 + 0.30 - 0.15 = 0.55, not 0.45.

Unit: probability-trees-and-conditional-expectations

Question 2Exam level

A stock has three possible year-end prices: $50 with probability 0.20, $60 with probability 0.50, and $75 with probability 0.30. The current price is $60. The expected return is closest to:

How sure are you?

Correct: B. The correct answer is 4.17%.
A. The $60 state has the highest probability and $60 is also the current price, so zero return feels anchored. Expected value weights all outcomes, not just the modal one.
C. You might compute unweighted average: (-16.67 + 0 + 25) / 3 = 2.78% (arithmetic), or incorrectly assign equal weights. Expected value uses probability weights, not equal weights.

Unit: probability-trees-and-conditional-expectations

Question 3Exam level

A portfolio has two assets. Asset A has expected return 8% and weight 60%. Asset B has expected return 12% and weight 40%. The variance of Asset A is 0.0100, variance of Asset B is 0.0225, and the covariance between A and B is 0.0080. The variance of the portfolio is closest to:

How sure are you?

Correct: B. The correct answer is 0.01104.
A. You might use weighted average of variances only: (0.60)(0.0100) + (0.40)(0.0225) = 0.006 + 0.009 = 0.015. Not this number, but similar errors drop the covariance term entirely. The 2·wA·wB·Cov term accounts for how assets move together. Omitting it ignores the diversification effect and always underestimates or misestimates variance.
C. You might forget to square the weights: use wA·σA² + wB·σB² + 2·Cov instead of wA²·σA² + wB²·σB². Weights must be squared in the variance formula; they are not squared in the expected return formula. This is the most common formula-confusion trap.

Unit: probability-trees-and-conditional-expectations

Question 4Exam level

An investment manager has historically outperformed in bull markets 70% of the time. The probability of a bull market in any given year is 60%. Given that the manager outperformed last year, what is the probability that it was most likely a bull market? [P(bull) = 0.60, P(outperform|bull) = 0.70, P(outperform|bear) = 0.20]

How sure are you?

Correct: B. The correct answer is 84%.
A. P(outperform|bull) = 70% is given in the problem. You might confuse P(A|B) with P(B|A) and just report the given conditional probability in the reverse direction. P(bull|outperform) ≠ P(outperform|bull). This is the Bayes direction-reversal trap. The exam is explicitly testing whether candidates understand that inverting the conditioning changes the answer.
C. 42% = P(outperform|bull) × P(bull) = 0.70 × 0.60, which is the Bayes numerator. You might compute the numerator but forget to divide by P(outperform). The numerator alone is a joint probability P(outperform AND bull), not the conditional probability P(bull|outperform). Division by the marginal is mandatory.

Unit: probability-trees-and-conditional-expectations

Question 5Exam level

Events A and B are mutually exclusive. P(A) = 0.35, P(B) = 0.45. P(A or B) is closest to:

How sure are you?

Correct: B. The correct answer is 0.80.
A. You might multiply: P(A) × P(B) = 0.35 × 0.45, thinking 'and' means multiply. This conflates the multiplication rule (for joint probability of independent events) with the question asking for 'or'. Mutually exclusive means they cannot both occur simultaneously, so P(A and B) = 0. The 'or' question uses the addition rule.
C. You might be tempted to choose 0.2025 if you mistakenly multiplied the probabilities of A and B, but this violates the rule for mutually exclusive events where you should add the probabilities, not multiply them, leading to the correct sum of 0.80.

Unit: probability-trees-and-conditional-expectations

Question 6Exam level

Events A and B are independent. P(A) = 0.40, P(B) = 0.30. P(A|B) is closest to:

How sure are you?

Correct: A. The correct answer is 0.40.
B. You might compute P(A) × P(B) = 0.12, which is P(A and B) for independent events, not P(A|B). P(A and B) = P(A) × P(B) for independent events, but the question asks for P(A|B), which equals P(A) itself.
C. You might think conditional probability always requires additional information about the joint event. The definition of independence precisely provides this: P(A|B) = P(A) whenever A and B are independent. This is not just a convenience. It IS the definition of independence.

Unit: probability-trees-and-conditional-expectations

Question 7Exam level

A portfolio manager believes the economy has a 60% probability of expansion and a 40% probability of recession. During expansion, portfolio return is expected to be 15%. During recession, expected return is -5%. The expected portfolio return is closest to:

How sure are you?

Correct: B. The correct answer is 7%.
A. You might average 15% and -5% equally: (15 - 5)/2 = 5%, ignoring probability weights. The scenario has asymmetric probabilities (60% vs 40%). Equal weighting is only valid when probabilities are equal.
C. You might use only the expansion return and discount it slightly without using the formula. All scenarios with non-zero probability must be included in the expected value calculation.

Unit: probability-trees-and-conditional-expectations

Question 8Exam level

A committee of 3 people is to be selected from a group of 8 candidates. The number of different committees possible is closest to:

How sure are you?

Correct: B. The correct answer is 56.
A. You might compute the permutation P(8,3) = 8!/(8-3)! = 8 × 7 × 6 = 336 instead of the combination. They include order when order does not matter. A committee has no order. {A, B, C} and {B, A, C} are the same committee. Dividing by r! (= 3! = 6) removes the overcounting: 336/6 = 56.
C. You might compute 8³ = 512, or use 2^8 = 256 (thinking of binary selection per person), reflecting confusion about when to use factorial methods. Neither exponentiation formula applies here. The combination formula applies when selecting r items from n without replacement and without regard to order.

Unit: probability-trees-and-conditional-expectations

Question 9Exam level

An analyst needs to assign 10 employees to 3 departments: 4 to Research, 3 to Trading, and 3 to Compliance. The number of different assignment schemes possible is closest to:

How sure are you?

Correct: B. The correct answer is 4,200.
A. You might compute 10! = 3,628,800 and forget to divide by the group-size factorials. Within each department, the assignment order does not matter. The labeling formula divides out the within-group permutations.
C. You might compute C(10,4) = 210, getting the first group selection but not accounting for distributing the remaining 6 employees. You must account for all three groups simultaneously. C(10,4) only selects the Research group; the full labeling formula handles all groups at once.

Unit: probability-trees-and-conditional-expectations

Question 10Exam level

The covariance between two assets is 0.0060. The standard deviation of Asset 1 is 0.10 and the standard deviation of Asset 2 is 0.15. The correlation coefficient is closest to:

How sure are you?

Correct: B. The correct answer is 0.40.
A. You might divide by the sum of standard deviations rather than the product: 0.0060 / (0.10 + 0.15) = 0.024 ≈ (arithmetic variant of this error). The denominator is σ1 × σ2 (product), not σ1 + σ2 (sum). Correlation is defined as covariance scaled by the product of standard deviations.
C. You might divide the covariance by one standard deviation and forget to divide by the second, or mix up variance and standard deviation. Correlation is bounded between -1 and +1. Any answer outside that range signals a formula error. Seeing 4.00 should immediately trigger a recalculation.

Unit: probability-trees-and-conditional-expectations

Question 11Exam level

Using the data from Question 7 (P(expansion) = 0.60, E(R|expansion) = 15%, P(recession) = 0.40, E(R|recession) = -5%, E(R) = 7%), the variance of portfolio returns is closest to:

How sure are you?

Correct: B. The correct answer is 0.0096.
A. You might compute only (0.15 - 0.07)² = 0.0064 from the expansion scenario and stops. Variance sums across all scenarios weighted by probability. One scenario alone is not the variance.
C. You might compute the unweighted average of squared deviations: (0.0064 + 0.0144) / 2 = 0.0104, or adds terms without weighting. Probability weights must be applied. Equal weights are wrong when scenario probabilities are unequal.

Unit: probability-trees-and-conditional-expectations

Question 12Harder

An equity analyst states that two events. 'the stock price rises' and 'the company beats earnings'. Are mutually exclusive. A colleague says they are independent. Which is most likely correct, and why?

How sure are you?

Correct: B. The correct answer is Neither; if they were mutually exclusive they could not be independent, and vice versa.
A. The word 'exclusive' suggests separation, which candidates conflate with independence. Mutually exclusive events can (and often do) have a causal relationship. Knowing one occurred tells you the other did not.
C. Candidates who don't know the formal definitions think the terms can coexist. For any two events A and B where P(A) > 0 and P(B) > 0: if they are mutually exclusive, P(A and B) = 0, so P(A|B) = P(A and B)/P(B) = 0 ≠ P(A). Therefore mutually exclusive events with positive individual probabilities are never independent.

Unit: probability-trees-and-conditional-expectations

Question 13Exam level

Five finalists are being ranked 1st, 2nd, and 3rd for an investment banking position. The number of different rankings possible is closest to:

How sure are you?

Correct: B. The correct answer is 60.
A. You might compute C(5,3) = 10. Using combination when order matters. Rankings are ordered. 1st: Alice, 2nd: Bob is a different outcome from 1st: Bob, 2nd: Alice. Use permutation P(5,3) = 60.
C. You might compute 5! = 120, assuming all 5 are ranked into all 5 positions. Only 3 of the 5 positions are being filled. P(5,3) = 5!/2! = 60, not 5! = 120.

Unit: probability-trees-and-conditional-expectations