Question 5

If a story has at least 100 likes, compute the probability it is fake news.

Let:

• F be the event the story is fake news.
• R be the event the story is real news.
• L be the event the story has at least 100 likes.

We are given:

• P(F) = 0.3
• P(R) = 0.7
• P(L|F) = 0.8
• P(L|R) = 0.08

We want P(F|L), which by Bayes’ Theorem is:

P(F|L) = [P(L|F) * P(F)] / [P(L|F)*P(F) + P(L|R)*P(R)]
= (0.8 * 0.3) / (0.8 * 0.3 + 0.08 * 0.7)
= 0.24 / (0.24 + 0.056)
= 0.24 / 0.296
= 60 / 74
= 30 / 37

Answer:
Numerator: 30
Denominator: 37

Question 6

You roll a fair 6-sided die two times and get paid the higher of the two rolls in dollars if the rolls are different. If they are the same, you get paid $0.

There are 36 total outcomes.

There are 6 outcomes where both rolls are the same: (1,1), (2,2), ..., (6,6). In these, the payoff is $0.

For the remaining 30 outcomes, the payoff is the maximum of the two dice. We compute the sum of those:

Let’s compute the sum of the maximums of all 30 different-roll outcomes:

Pair Type	Number of Outcomes	Max Value	Total Contribution
(1,2),(2,1)	2	2	4
(1,3),(3,1)	2	3	6
(1,4),(4,1)	2	4	8
(1,5),(5,1)	2	5	10
(1,6),(6,1)	2	6	12
(2,3),(3,2)	2	3	6
(2,4),(4,2)	2	4	8
(2,5),(5,2)	2	5	10
(2,6),(6,2)	2	6	12
(3,4),(4,3)	2	4	8
(3,5),(5,3)	2	5	10
(3,6),(6,3)	2	6	12
(4,5),(5,4)	2	5	10
(4,6),(6,4)	2	6	12
(5,6),(6,5)	2	6	12

Sum of contributions = 4 + 6 + 8 + 10 + 12 + 6 + 8 + 10 + 12 + 8 + 10 + 12 + 10 + 12 + 12 = 140

Expected value = 140 / 36 = 3.888...

Rounded to the nearest cent: $3.89

Answer:
Expected Payoff: $3.89

Question 7

There are 5 people in line. Each has a 30% chance of choosing a muffin. We want the probability that at most 2 people want a muffin.

Let X be the number of people who want a muffin. X ~ Binomial(n = 5, p = 0.3)

We want P(X ≤ 2):

P(X=0) = C(5,0) * (0.3)^0 * (0.7)^5 = 1 * 1 * 0.16807 = 0.16807
P(X=1) = C(5,1) * (0.3)^1 * (0.7)^4 = 5 * 0.3 * 0.2401 = 0.36015
P(X=2) = C(5,2) * (0.3)^2 * (0.7)^3 = 10 * 0.09 * 0.343 = 0.3087

Total = 0.16807 + 0.36015 + 0.3087 = 0.83692

Rounded to the nearest whole percent: 84%

Answer:
Probability: 84%

Question 8

Let A, B, and C be the number of cats Asta, Bronya, and Clara have, respectively.

Let’s denote the constraints:

ii. A = B * 1.3
iii. C = (A + B) / 2
iv. A ≥ C + 4

Try integer values:

Assume B = 10 → A = 13
Then C = (13 + 10)/2 = 11.5 → Invalid
Try B = 20 → A = 26 → C = 23 → A = 26 ≥ 27? → No
Try B = 30 → A = 39 → C = 34.5 → A = 39 ≥ 38.5 → Yes

Check who has the most cats: Asta (39), Clara (34.5), Bronya (30)

Now evaluate who lies and who tells the truth:

i. Bronya to Clara: “You have the most cats.” → Clara has more than Bronya → Bronya is speaking to someone with more cats → She lies → FALSE (correct)

ii. Asta to Bronya: “I have exactly 30% more cats than you.” → Asta has more → Speaking to someone with fewer cats → Tells the truth → TRUE

iii. Asta to Clara: “You have the average between me and Bronya.” → (39+30)/2 = 34.5 → Clara has 34.5 → Speaking to someone with fewer cats → Should tell truth → TRUE

iv. Clara to Asta: “You have at least 4 more than me.” → 39 - 34.5 = 4.5 ≥ 4 → Statement is TRUE. But Clara is speaking to someone with more cats → Should lie → FALSE

So only i and iv are false, which fits. Therefore:

Answer:
Number of cats Clara has: 34.5

But question says "all have different number of cats" and likely integers.

Try B = 20 → A = 26 → C = (26 + 20)/2 = 23 → A = 26 ≥ 27? No
Try B = 10 → A = 13 → C = 11.5 → Not integer

Eventually, trying B = 10 → A = 13 → C = 11.5 → Doesn't work

Try B = 20 → A = 26 → C = 23
A = 26 ≥ 27? No → Invalid
Try B = 30 → A = 39 → C = 34.5 → Again not integer

Try B = 40 → A = 52 → C = 46 → A = 52 ≥ 50 → Yes
Bronya to Clara (40 vs 46): she lies → FALSE
Asta to Bronya (52 vs 40): truth → 52 = 1.3 * 40 = TRUE
Asta to Clara (52 vs 46): 46 = (52 + 40)/2 = 46 → TRUE
Clara to Asta (46 vs 52): says Asta ≥ 4 more → TRUE but should lie → FALSE

So, only i and iv false → Acceptable

Answer:
Number of cats Clara has: 46

Question 9

River speed: 2 mph
Let p = paddling speed in still water (in mph)

Day 1:
Upstream for 4 hours → speed = p - 2
Distance = 4(p - 2)

Same day:
Downstream for 5 hours → speed = p + 2
Distance = 5(p + 2)

So:
4(p - 2) = 5(p + 2)
4p - 8 = 5p + 10
-8 - 10 = p
p = -18 → invalid

Wait. Distance must be the same both ways? No—it says they go upstream for 4 hours, then downstream for 5 hours, then the next day, they canoe back 23 miles upstream and arrive at 16:00.

So for return trip: They travel 23 miles upstream
Speed = p - 2
Time = 23 / (p - 2) hours
They arrive at 16:00

Let t = time they left
t + 23 / (p - 2) = 16:00

We need to find p using earlier journey:

From Day 1:
They paddle 4 hours upstream, then 5 hours downstream
Distance upstream = 4(p - 2)
Distance downstream = 5(p + 2)

Those distances are equal:

4(p - 2) = 5(p + 2)
4p - 8 = 5p + 10
-8 -10 = p
p = -18 → again invalid

Try again: They canoe 4 hours upstream → D1 = 4(p - 2)
Then turn around and canoe for 5 hours downstream → D2 = 5(p + 2)

So total downstream = D1 → same distance? Not unless the same point.

Maybe total distance from end point to starting point is 23 miles.

Let’s suppose 5(p + 2) = 23 miles

Then p = (23 / 5) - 2 = 4.6 - 2 = 2.6

Now on return trip, they canoe 23 miles upstream at p - 2 = 0.6 mph

Time = 23 / 0.6 = 38.33 hours → impossible

Try another:
Suppose p = 5 mph
Then downstream speed = 7 mph
Upstream speed = 3 mph
Downstream time = 5 hours → Distance = 35 miles
Upstream for 4 hours → Distance = 12 miles
So total trip = 12 miles up, 35 miles down → Difference = 23 miles

Now return trip is upstream 23 miles at 3 mph = 7.6667 hours

16:00 - 7.6667 hours ≈ 08:20

Answer:
Departure time: 08:20

Question 10

You baked 5 indistinguishable snickerdoodle cookies and 7 indistinguishable chocolate chip cookies. Compute the number of ways to arrange 6 of these cookies into a straight line.

We must choose how many of each cookie type to include, totaling 6 cookies. For each valid combination, count the permutations considering indistinguishability.

Valid combinations:

Snickerdoodles	Chocolate Chips	Ways
0	6	C(6,0) = 1
1	5	C(6,1) = 6
2	4	C(6,2) = 15
3	3	C(6,3) = 20
4	2	C(6,4) = 15
5	1	C(6,5) = 6

Total = 1 + 6 + 15 + 20 + 15 + 6 = 63

Answer:
Number of arrangements: 63

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