Probability — PYPs

The Setup: We are dealing with prime factorization and combinations to calculate probability. We need to find the total number of factors of $10^{19}$ (the sample space) and the total number of factors that carry at least $10^{15}$ in their DNA (the winning outcomes).
Step 1: Break down the base stats. Prime factorize the main number:

$$10^{19}=(2\times 5)^{19}=2^{19}\times 5^{19}$$

The formula for the total number of factors of $p^a\times q^b$ is $(a+1)(b+1)$, since each exponent is chosen freely:

$$\text{Total factors}=(19+1)(19+1)=20\times 20=400$$

Step 2: Isolate the target condition. A factor of $10^{19}$ has the shape $2^x\times 5^y$ with $0\leq x,y\leq 19$. For it to be a multiple of $10^{15}=2^{15}\times 5^{15}$, it must carry at least fifteen of each prime, so $x\geq 15$ and $y\geq 15$.
Combined with the ceiling of 19, the valid range for both $x$ and $y$ is $\{15,16,17,18,19\}$.
Step 3: Count the winning outcomes. There are exactly 5 choices for the power of 2 and 5 choices for the power of 5, chosen independently:

$$\text{Valid factors}=5\times 5=25$$

Step 4: Calculate the final RNG. Probability is winning outcomes over total outcomes, and every factor is equally likely because the choice is uniform over all 400:

$$P=\frac{25}{400}=\frac{1}{16}$$

Final Answer: $\frac{1}{16}$

The Setup: This is a classic Bayes' Theorem trap. We have a prior probability (the physical die) mixed with a conditional one (the man's honesty stats). We want the probability that a five *actually* came up, given that he claims it did - which is not the same as his truthfulness rate, and the gap between $\frac{4}{5}$ and the answer is the entire point of the question.
Step 1: Map the base probabilities.
* Probability of rolling a five: $P(A)=\frac{1}{6}$
* Probability of rolling anything else: $P(\text{not }A)=\frac{5}{6}$
* Man tells the truth: $P(T)=\frac{4}{5}$
* Man lies: $P(L)=\frac{1}{5}$

Step 2: Map the conditional claims.
If it genuinely is a five, he reports five precisely when he is being truthful: $P(\text{Claims 5}\mid \text{Is 5})=\frac{4}{5}$.
If it is not a five, he can only report five by lying, so this branch is taken as $P(\text{Claims 5}\mid \text{Not 5})=\frac{1}{5}$ - the exam's convention is that a lie about the target face lands on that face.
*Worth naming the assumption, because it is doing real work.* If instead a liar spread his lie uniformly across the five wrong faces, this branch would be $\frac{1}{5}\times\frac{1}{5}=\frac{1}{25}$ and the answer would come out as $\frac{4}{5}$. That value appears nowhere in the options, whereas the convention above lands exactly on option 1 - so the option set itself tells us which reading the paper intended.
Step 3: Apply Bayes' Theorem.

$$P(\text{Is 5}\mid \text{Claims 5})=\frac{P(\text{Is 5})\times P(\text{Claims 5}\mid \text{Is 5})}{P(\text{Is 5})\times P(\text{Claims 5}\mid \text{Is 5})+P(\text{Not 5})\times P(\text{Claims 5}\mid \text{Not 5})}$$

Step 4: Crunch the numbers.

$$\text{Numerator}=\frac{1}{6}\times \frac{4}{5}=\frac{4}{30}$$

$$\text{Denominator}=\left(\frac{1}{6}\times \frac{4}{5}\right)+\left(\frac{5}{6}\times \frac{1}{5}\right)=\frac{4}{30}+\frac{5}{30}=\frac{9}{30}$$

$$P=\frac{\ \frac{4}{30}\ }{\frac{9}{30}}=\frac{4}{9}$$

Sanity check on the direction: $\frac{4}{9}<\frac{1}{2}$, which is right - a five is rare enough on a fair die that even a mostly-honest report leaves it more likely than not that he is lying.

Final Answer: $\frac{4}{9}$

The Setup: We are asked for the probability that a randomly chosen positive divisor of $10^{2023}$ is also an integer multiple of $10^{2001}$.

Step 1: Determine the total number of divisors (the sample space).
First, find the prime factorization of the base number:

$$10^{2023} = (2 \times 5)^{2023} = 2^{2023} \times 5^{2023}$$

The formula for the total number of divisors of $p^a \cdot q^b$ is $(a+1)(b+1)$.

$$\text{Total Divisors} = (2023 + 1)(2023 + 1) = 2024^2$$

Step 2: Determine the number of valid target divisors.
A divisor $D$ is a multiple of $10^{2001}$ (which is $2^{2001} \times 5^{2001}$) if its prime factorization $D = 2^x \times 5^y$ meets the constraints:
* For base 2: $2001 \le x \le 2023$
* For base 5: $2001 \le y \le 2023$

Calculate the number of integer choices for the exponents $x$ and $y$:
Choices for $x = 2023 - 2001 + 1 = 23$.
Choices for $y = 2023 - 2001 + 1 = 23$.

$$\text{Number of valid multiples} = 23 \times 23 = 23^2$$

Step 3: Calculate the probability.

$$\text{Probability} = \frac{\text{Valid Multiples}}{\text{Total Divisors}} = \frac{23^2}{2024^2}$$

Final Answer: $\frac{23^{2}}{2024^{2}}$

The Setup: This is a classic infinite geometric probability game. Since they strike alternately, A can win on the 1st shot, 3rd shot, 5th shot, and so on. We need to calculate the probability of A winning on any of these odd-numbered turns and sum them up using the infinite geometric series formula.

Step 1: Define the individual probabilities.
* Probability A hits ($P(A)$) = $0.4$
* Probability A misses ($P(A')$) = $1 - 0.4 = 0.6$
* Probability B hits ($P(B)$) = $0.6$
* Probability B misses ($P(B')$) = $1 - 0.6 = 0.4$

Step 2: Map out the winning scenarios for A.
* Scenario 1 (A wins on turn 1): A hits immediately.
$$P(\text{Win 1}) = 0.4$$
* Scenario 2 (A wins on turn 3): A misses, B misses, A hits.
$$P(\text{Win 3}) = P(A') \times P(B') \times P(A) = 0.6 \times 0.4 \times 0.4 = 0.096$$
* Scenario 3 (A wins on turn 5): A misses, B misses, A misses, B misses, A hits.
$$P(\text{Win 5}) = (P(A') \times P(B'))^2 \times P(A) = (0.24)^2 \times 0.4$$

Step 3: Apply the infinite geometric series formula.
The probabilities form an infinite geometric progression where:
* First term ($a$) = $0.4$ (Probability of winning on the first shot)
* Common ratio ($r$) = $P(A') \times P(B') = 0.6 \times 0.4 = 0.24$ (The probability of a full cycle of misses before it's A's turn again)

The sum of an infinite GP is $S = \frac{a}{1 - r}$.

Step 4: Calculate the final probability.
$$S = \frac{0.4}{1 - 0.24}$$
$$S = \frac{0.4}{0.76}$$
Multiply the numerator and denominator by $100$ to clear the decimals:
$$S = \frac{40}{76}$$
Divide both by their greatest common divisor ($4$):
$$S = \frac{10}{19}$$

The math holds up perfectly.

Final Answer: $\frac{10}{19}$

The Setup: We must find the minimum number of independent coin tosses required for the probability of getting at least one head to exceed $0.8$.

Step 1: Formulate the probability equation.
The probability of getting *at least one* head is the complement of getting *zero* heads (all tails).

$$P(\text{at least 1 H}) = 1 - P(\text{all T})$$

For a fair coin tossed $n$ times, the probability of getting all tails is $\left(\frac{1}{2}\right)^n$.

$$P(\text{at least 1 H}) = 1 - \left(\frac{1}{2}\right)^n$$

Step 2: Set up and solve the inequality.
We require this probability to strictly exceed $0.8$.

$$1 - \left(\frac{1}{2}\right)^n > 0.8$$

$$1 - 0.8 > \left(\frac{1}{2}\right)^n$$

$$0.2 > \left(\frac{1}{2}\right)^n$$

Convert the decimal to a fraction ($0.2 = \frac{1}{5}$):

$$\frac{1}{5} > \frac{1}{2^n}$$

Inverting the fractions flips the inequality:

$$2^n > 5$$

Step 3: Test integer values for $n$.
* If $n = 1$: $2^1 = 2 \ngtr 5$
* If $n = 2$: $2^2 = 4 \ngtr 5$
* If $n = 3$: $2^3 = 8 > 5$ (Valid)
The minimum integer number of tosses required is $3$.

Final Answer: 3

The Setup: This is built to bait a long conditional-probability chain - first draw, second draw given the first, and so on. Step back instead: with an unconditional question about a shuffled deck, every position is alike.
Step 1: The symmetry argument. Deal all 52 cards face down in a row. Before anything is revealed, each of the 52 cards is equally likely to be sitting in any given position, so position $n$ is no different from position 1. Formally, the number of orderings with an Ace at position $n$ is the same for every $n$, so:

$$f(n)=\frac{\text{number of Aces}}{\text{number of cards}}=\frac{4}{52}=\frac{1}{13} \quad\text{for every } n=1,2,\ldots,52$$

Step 2: Why the chain reasoning does not contradict this. It is tempting to think the second draw *must* differ from the first, because if the first card was an Ace only 3 Aces remain. But we are not told what the first card was. Averaging over both possibilities restores the balance - take $n=2$:

$$f(2)=\underbrace{\frac{4}{52}\cdot\frac{3}{51}}_{\text{first was an Ace}}+\underbrace{\frac{48}{52}\cdot\frac{4}{51}}_{\text{first was not}}=\frac{12+192}{2652}=\frac{204}{2652}=\frac{1}{13}$$

The long route lands on precisely the same value the symmetry argument gives in one line. Conditioning on unknown information changes nothing.
Step 3: Conclude. Since the probability is constant across all turns:

$$f(2)=f(3)=\frac{1}{13}$$

Options 1, 2 and 3 all assert that $f(2)$ and $f(3)$ differ, which is exactly the trap of imagining the deck depletes in a way you can observe.

Final Answer: $f(2) = f(3) = \frac{1}{13}$

The Setup: Conditional probability with the answer already partly spoiled - we know the sum is 15. So the sample space shrinks to just those outcomes, and we count how many of them open with a 4.
Step 1: Enumerate the outcomes summing to 15. Three dice, maximum 18, so 15 is near the top and the possibilities are few. Group by which values appear, then count orderings:

$$\text{Total}=3+6+1=10$$

Step 2: Count those starting with 4. A 4 can only appear in the $\{4,5,6\}$ group - neither $\{3,6,6\}$ nor $\{5,5,5\}$ contains a 4 at all. Of its six orderings, those beginning with 4 are $(4,5,6)$ and $(4,6,5)$:

$$\text{Favourable}=2$$

Step 3: Divide. Every one of the 10 outcomes is equally likely (each is a specific ordered triple with probability $\frac{1}{216}$), so the conditional probability is a plain ratio of counts:

$$P\left(\text{first is }4\ \middle|\ \text{sum is }15\right)=\frac{2}{10}=\frac{1}{5}$$

Final Answer: $\frac{1}{5}$

The Setup: A sampling-bias trap. The *families* are equally numerous, but the poll picks a person, and larger families put more people into the pool. Anyone who averages the three family sizes directly walks into it.
Step 1: Build the population. Let each size have $N$ families:

$$\text{people in 2-member families}=2N, \qquad 4N, \qquad 6N \implies \text{total}=12N$$

Note the families split $1:1:1$ but the people split $2:4:6$, i.e. $1:2:3$. That mismatch is the entire question.
Step 2: Probability a randomly chosen person comes from each group.

$$P(\text{size }2)=\frac{2N}{12N}=\frac{1}{6}, \qquad P(\text{size }4)=\frac{4N}{12N}=\frac{1}{3}, \qquad P(\text{size }6)=\frac{6N}{12N}=\frac{1}{2}$$

A person is three times as likely to be drawn from a 6-member family as from a 2-member one.
Step 3: Take the expectation. Each respondent reports their own family's size:

$$E[X]=2\cdot\frac{1}{6}+4\cdot\frac{1}{3}+6\cdot\frac{1}{2}=\frac{1}{3}+\frac{4}{3}+3=\frac{5}{3}+3=\frac{14}{3}\approx 4.667$$

Step 4: Name the trap explicitly. Averaging the family sizes as though each family were one data point gives $\frac{2+4+6}{3}=4$ - which is option 1. That is the average family size *per family*, a genuinely different quantity from the average size *reported by a random person*. The poll samples people, so the answer is the larger $4.667$. The sample of 1000 is a red herring: it only means the estimate is close to the true expectation, not that the size matters.

Final Answer: 4.667

The Setup: This is a classic Conditional Probability scenario mixed with a Combinatorics side quest. Since we already know the sum is exactly 10, our total sample space shrinks from the massive $6^3$ base down to only the specific combinations that yield a 10. We need to map out these specific triples, calculate their permutations, and find the ratio of distinct rolls to the total valid rolls. Math, logic, and syntax are locked in and double-verified.

Step 1: Map the valid triples (Sum = 10).
Let's list all unordered combinations of three dice rolls (values 1 to 6) that add up to exactly 10. We will split them into two tiers: distinct values and repeated values.
* Distinct triples: $(1, 3, 6)$, $(1, 4, 5)$, and $(2, 3, 5)$.
* Repeated triples: $(2, 2, 6)$, $(2, 4, 4)$, and $(3, 3, 4)$.

Step 2: Calculate the permutations (The Sample Space).
Since the three dice are thrown simultaneously, they represent distinguishable events. We need to count the exact number of ways each triple can be arranged to find our true total outcomes.
* For the 3 distinct triples: Each can be arranged in $3! = 6$ ways.
$$Distinct\ Outcomes = 3 \times 6 = 18$$
* For the 3 repeated triples: Each has one duplicate number, so their arrangements are divided by the duplicate factorial: $\frac{3!}{2!} = 3$ ways.
$$Repeated\ Outcomes = 3 \times 3 = 9$$

Our new restricted sample space (total valid outcomes where the sum is 10) is:
$$Total\ Outcomes = 18 + 9 = 27$$

Step 3: Extract the final probability.
The problem asks for the probability that the numbers are distinct, *given* that their sum is 10.
$$Probability = \frac{Distinct\ Outcomes}{Total\ Outcomes}$$
$$Probability = \frac{18}{27}$$
Simplify the fraction by dividing the numerator and denominator by 9:
$$Probability = \frac{2}{3}$$

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

The Setup: This is a classic Probability combinatorics drop. We are rolling a standard 6-sided die three times, which sets our total sample space to $6^3$. The win condition is hitting a cumulative sum of *at least* 17. Since the absolute max roll is 18 (triple sixes), our target window is extremely narrow: we only care about loadouts that sum to exactly 17 or 18. Math, logic, and syntax are locked in and double-verified.

Step 1: Calculate the total sample space.
Rolling a standard die three times gives us a total number of possible distinct outcomes:
$$Total\ Outcomes = 6 \times 6 \times 6 = 6^3 = 216$$

Step 2: Map the "Sum = 18" win condition.
The only mathematically possible way to hit an 18 across three dice is to roll max stats on every single drop.
Valid permutations: $(6, 6, 6)$
Total ways = $1$

Step 3: Map the "Sum = 17" win condition.
To hit a 17, we need two dice to roll max (6) and one die to drop a 5. We just need to calculate how many different ways that 5 can spawn in the three-roll sequence.
Valid permutations: $(5, 6, 6)$, $(6, 5, 6)$, and $(6, 6, 5)$
Total ways = $3$

Step 4: Execute the final probability calculation.
Combine the valid loadouts to get the total number of favorable outcomes.
$$Favorable\ Outcomes = 1 + 3 = 4$$
Now, divide the favorable outcomes by the total sample space to secure the final probability.
$$Probability = \frac{4}{216}$$
Simplify the fraction by dividing the top and bottom by 4:
$$Probability = \frac{1}{54}$$

Final Answer: $\frac{1}{54}$