Use the complement rule: calculate the probability that the second roll is the same as the first, then subtract from 1. Alternatively, condition on the first roll and count favorable outcomes for the second roll.
Pro tip: State your assumptions clearly (fair die, independent rolls) and consider mentioning both the complement and direct counting methods to show versatility. For trading firms like Optiver, emphasize quick mental math and logical reasoning.
Clarify that we roll a fair six-sided die twice, and we want the probability that the second roll differs from the first. Assume independence and fairness.
Decide between using the complement rule (1 - P(same)) or direct counting (condition on first roll). Both are valid; pick the one you find quicker.
If using complement: The probability the second roll equals the first is 1/6, since there is exactly one matching outcome out of six equally likely outcomes.
Subtract from 1: 1 - 1/6 = 5/6. If using direct counting: After any first roll, 5 of the 6 possible second rolls are different, so probability is 5/6.
Check that the answer makes sense (e.g., probability should be high since only one match is possible). Present the answer clearly, stating the final probability as 5/6 or approximately 83.33%.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Recognize that discarding cards face-down without looking does not change the probability for any remaining card. The 11th card is equally likely to be any of the 52 cards, so the probability it is red is simply the proportion of red cards in the deck. State the answer clearly and explain the symmetry argument.
Pro tip: Emphasize that the discarding process is irrelevant because the cards are not observed; this demonstrates understanding of conditional probability and symmetry. Avoid overcomplicating with conditional cases unless asked to elaborate.
Identify that we have a standard 52-card deck with 26 red and 26 black cards. We discard 10 cards face-down without looking, then ask for the probability that the 11th card is red.
Since the discarded cards are not observed, the remaining deck is still a random permutation of all 52 cards. Thus, the 11th card is equally likely to be any of the 52 cards.
The probability that a randomly chosen card is red is the number of red cards divided by total cards: 26/52 = 1/2.
Clearly state that the probability is 1/2 or 50%, and briefly explain why the discarding step does not affect the result.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Clarify that the dice are fair and independent, then compute the total number of equally likely outcomes (36). Count the favorable outcomes for sum 11 (2 ways) and sum 12 (1 way), add them, and divide by 36 to get the probability.
Pro tip: State the answer as a simplified fraction (1/12) and mention that the events are mutually exclusive, so you add probabilities. This shows you understand both the calculation and the underlying probability rules.
Each die has 6 faces, so there are 6 × 6 = 36 equally likely outcomes when rolling two fair dice.
List the pairs that sum to 11: (5,6) and (6,5). That gives 2 favorable outcomes.
List the pairs that sum to 12: (6,6). That gives 1 favorable outcome.
Since the events are mutually exclusive, add the counts: 2 + 1 = 3 favorable outcomes.
Divide favorable outcomes by total outcomes: 3/36 = 1/12. State the final answer clearly.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Two favorable sequences out of 8 total: HHH and TTT.
Clarify that the coin is fair and flips are independent, then compute the probability of all heads and all tails separately, and sum them. Alternatively, count the favorable outcomes (HHH, TTT) out of 8 equally likely outcomes. State the final probability as 1/4.
Pro tip: Show that you can generalize: for n flips, the probability is 2*(1/2)^n = (1/2)^(n-1). This demonstrates pattern recognition and mathematical maturity, which is valued in quantitative roles.
Confirm that the coin is fair (probability 1/2 for heads or tails) and that flips are independent. This ensures a common understanding before calculating.
List all possible outcomes of 3 flips: there are 2^3 = 8 equally likely sequences (HHH, HHT, HTH, HTT, THH, THT, TTH, TTT).
Determine which outcomes satisfy 'all three flips show the same result': HHH and TTT. So there are 2 favorable outcomes.
Divide the number of favorable outcomes by the total number of outcomes: 2/8 = 1/4. Alternatively, compute P(all heads) + P(all tails) = (1/2)^3 + (1/2)^3 = 1/8 + 1/8 = 1/4.
Conclude that the probability is 1/4 or 25%. Optionally, mention the general formula for n flips: 2*(1/2)^n = (1/2)^(n-1).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
I spent a moment trying to set up a combinatorics calculation before realizing this is just pigeonhole.
Use complementary counting: calculate the probability that no box has more than 4 coins, then subtract from 1. The total number of ways to distribute 61 identical coins into 15 distinct boxes is C(75,14). The favorable ways are the number of solutions to x1+...+x15=61 with each xi ≤ 4, which can be found using inclusion-exclusion or generating functions.
Pro tip: In probability problems with 'at least one', always consider the complement first. Also, clarify whether coins are identical or distinct; for this question, identical coins lead to a cleaner combinatorial solution.
The event 'at least one box has more than 4 coins' is easier to handle via its complement: 'no box has more than 4 coins' (i.e., each box has at most 4 coins).
Assuming identical coins and distinct boxes, the total number of ways to distribute 61 coins into 15 boxes is the number of nonnegative integer solutions to x1+...+x15=61, which is C(61+15-1,15-1)=C(75,14).
Count the number of solutions to x1+...+x15=61 with 0 ≤ xi ≤ 4 for all i. This can be done using inclusion-exclusion or generating functions: coefficient of x^61 in (1+x+...+x^4)^15.
Divide the favorable count by the total count to get P(no box >4). Then the desired probability is 1 minus this value.
If possible, simplify the expression or provide a numerical approximation. State the final probability clearly.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.