Start by defining the sample space of 36 equally likely outcomes for two dice. Then, for each possible sum from 2 to 12, count the number of outcomes that yield that sum and divide by 36 to get the probability. Present the distribution clearly, perhaps in a table or as a formula.
Pro tip: Mention that the distribution is symmetric around 7 and that the probabilities follow a triangular pattern, which can be derived combinatorially. This shows deeper insight and can lead to a discussion of generating functions or convolution.
State that each die has 6 faces, so there are 6 x 6 = 36 equally likely outcomes when rolling two dice.
Identify that the sum can range from 2 (1+1) to 12 (6+6).
For each sum s, count the number of pairs (i, j) with i + j = s, where i and j are integers from 1 to 6. This can be done systematically or by using the formula: count = 6 - |s - 7| for s from 2 to 12.
Divide each count by 36 to get the probability for each sum. For example, P(sum=2) = 1/36, P(sum=7) = 6/36 = 1/6.
Summarize the probabilities in a table or list, and note the symmetry and the most likely sum (7).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
This is the one that trips people up and I almost let it trip me too.
Recognize that discarding unseen cards does not change the overall probability of drawing a red card from the remaining deck. The probability remains the same as drawing a red card from a full deck: 26 red out of 52, i.e., 1/2. Explain that the unseen discarded cards are equally likely to be red or black, so the remaining deck's composition is symmetric.
Pro tip: Emphasize that this is a classic probability puzzle testing intuition about conditional probability and symmetry. Avoid overcomplicating with hypergeometric calculations; the key is that without information, the order of drawing doesn't matter.
Clarify that 10 cards are removed face-down without being seen, and we want the probability that the next card drawn is red.
Note that the deck is symmetric with respect to red and black, and the discarded cards are unknown, so they don't bias the remaining deck.
Use the fact that the probability of the next card being red is the same as the probability that any specific card in the deck is red, which is 26/52 = 1/2.
Consider that if you randomly permute the deck, the 11th card is equally likely to be any of the 52 cards, so the probability it is red is 1/2.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
First, clarify that the dice are fair and independent, then enumerate all 36 equally likely outcomes. Count the outcomes that sum to 11 or 12, and compute the probability as the ratio of favorable outcomes to total outcomes.
Pro tip: State your assumptions upfront (fair dice, independent rolls) and consider mentioning how you would verify the result with a quick simulation or by generalizing to n dice, showing both analytical and computational thinking.
Confirm that the dice are fair and independent, and that each die has 6 equally likely faces.
List all possible outcomes as ordered pairs (die1, die2), totaling 36 equally likely outcomes.
Count the outcomes that sum to 11: (5,6) and (6,5); and sum to 12: (6,6). Total favorable outcomes = 3.
Divide the number of favorable outcomes by the total number of outcomes: 3/36 = 1/12.
Optionally, verify by simulation or mention how to extend the approach to other sums or multiple dice.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Recognize this as a classic gambler's ruin problem with a fair coin, where the probability of reaching the goal from a starting bankroll is simply the ratio of the starting bankroll to the goal. State the formula P(reach N | start i) = i/N for a fair game, then plug in i=10 and N=20 to get 1/2. Optionally, briefly justify using the martingale property or solving the recurrence relation.
Pro tip: Mention that this linear result is a special case of the fair game and that for biased games the formula becomes more complex; this shows you understand the underlying assumptions and can generalize.
Recognize that this is a gambler's ruin problem with absorbing barriers at 0 and 20, and a fair coin (p=0.5).
For a fair game, the probability of reaching N before 0 starting from i is i/N. This can be derived using martingales or by solving the recurrence P_i = 0.5 P_{i-1} + 0.5 P_{i+1} with boundary conditions P_0=0, P_N=1.
Plug in i=10 and N=20 to get P = 10/20 = 1/2.
Confirm that the answer makes intuitive sense: starting halfway to the goal in a fair game gives a 50% chance. Optionally, mention that the expected duration is i(N-i) = 100 steps.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Binomial distribution with p = 1/6 and n = N trials.
Model each roll as an independent Bernoulli trial where success is the specific face appearing with probability 1/6. Then the count of successes in N rolls follows a Binomial(N, 1/6) distribution, so the expected value is N/6 and the variance is N*(1/6)*(5/6) = 5N/36. Clearly state the distribution and derive the results.
Pro tip: Mention that this is a classic binomial setup and that the independence of rolls is crucial. Also, note that for large N, the binomial can be approximated by a normal distribution, which is often useful in trading contexts.
Let X be the number of times the specific face appears in N rolls. Each roll is independent and has probability p = 1/6 of showing that face.
Since there are N independent trials with constant success probability, X follows a Binomial distribution: X ~ Binomial(N, p).
For a binomial random variable, E[X] = N * p. Substitute p = 1/6 to get E[X] = N/6.
For a binomial random variable, Var(X) = N * p * (1-p). Substitute p = 1/6 to get Var(X) = N * (1/6) * (5/6) = 5N/36.
State the final answers: expected count = N/6, variance = 5N/36. Optionally, mention that the standard deviation is sqrt(5N/36).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
For the product to be odd, every single die has to show an odd number.
Recognize that the product is odd only if all four dice show odd numbers. Compute the probability for one die to be odd (3/6 = 1/2) and raise it to the fourth power, since the rolls are independent. The final answer is (1/2)^4 = 1/16.
Pro tip: State the independence assumption explicitly and note that the number of odd outcomes per die is 3, so the probability is (3/6)^4. This shows you understand the underlying symmetry and avoids overcomplicating the problem.
The product of four integers is odd if and only if every integer is odd. So all four dice must show an odd number (1, 3, or 5).
For a fair six-sided die, there are 3 odd outcomes out of 6, so the probability of rolling an odd number is 3/6 = 1/2.
Since the dice rolls are independent, the probability that all four are odd is the product of the individual probabilities: (1/2) × (1/2) × (1/2) × (1/2) = (1/2)^4.
Calculate (1/2)^4 = 1/16. So the probability is 1/16, or 6.25%.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.