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Each of 20 families selected to take part in a treasure hunt consist of a mother, father, son, and daughter. Assuming that they look for the treasure in pairs that are randomly chosen from the 80 participating individuals and that each pair has the same probability of finding the treasure, calculate the probability that the pair that finds the treasure includes a mother but not her daughter.

Solution I thought

Sample space: $80\times79$
As there are 20 mothers so we select and each have one daughter so event will have $20\times78$ samples. This gives answer $0.247$.
But answer in book is $0.3734$.

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    $\begingroup$ I'm not sure how Ross gets this answer. There are $\binom{80}{2} = 3160$ possible pairs (not $80 \times 79 = 6320$), and $0.3734 \times 3160 \approx 1180$, so evidently there are $1180$ qualifying pairs. There are $20 \times 40 = 800$ pairs that consist of a mother and any male, and $20 \times 19 = 380$ pairs that consist of a mother and a daughter not her own, so that could be the $1180$. But then it seems there are $\binom{20}{2} = 190$ additional pairs that consist of two mothers. I must be misunderstanding the question (or I made a mathematics error), so I'll wait to see what others say. $\endgroup$
    – Brian Tung
    Apr 12 at 6:36
  • $\begingroup$ Is this really from Ross? I'm flipping through it and I don't find any Question 18 that refers to a treasure hunt. $\endgroup$ May 1 at 19:06
  • $\begingroup$ @MatthewLeingang See chapter 2 "Axioms of Probability" edition 10. There may be a difference due to the continent. Mine is a Indian. $\endgroup$ May 2 at 11:54
  • $\begingroup$ @AbhishekSingh I see. I have an older edition. $\endgroup$ May 2 at 12:28

2 Answers 2

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Consider a randomly chosen pair which find the treasure. Let $A_1$ be the event that the first person in the pair is a mother; let $A_2$ be the event that the second person in the pair is a mother. Let $B_1$ be the event that the second person in the pair is not a daughter of the first person in the pair (and similarly define $B_2$). Then $$ \mathbb{P}(A_1)=\frac{20}{80}, \quad \mathbb{P}(B_1\mid A_1)=\frac{78}{79}, $$ as there are 20 mothers to choose from 80 people, and after that, all but 1 amongst the remaining people are not her daughter. The same equalities hold for $A_2$ and $B_2$. As a result, the required quantity is $$ \mathbb{P}(A_1 B_1 \text{ or }A_2 B_2 )=\mathbb{P}(A_1 B_1)+\mathbb{P}(A_2 B_2)-\mathbb{P}(A_1 B_1 A_2 B_2). $$ Now $$ \mathbb{P}(A_1 B_1 A_2 B_2)=\mathbb{P}(A_1 A_2)=\mathbb{P}(A_2 \mid A_1)\mathbb{P}(A_1)=\frac{19}{79}\,\frac{20}{80} $$ as once the first person is chosen to be a mother, there are 19 mothers to choose from. Hence the answer is $$ \frac{78}{79}\,\frac{20}{80}+\frac{78}{79}\,\frac{20}{80}-\frac{19}{79}\,\frac{20}{80}=\frac{137}{316}\approx 0.4335. $$

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    $\begingroup$ Your reasoning right but the answer given in book is 0.3734. I am waiting for few more comments or answers if everyone gets same answer then will be ensured that books answer is wrong. $\endgroup$ Apr 12 at 7:18
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I think the point is in having a mother, meaning the outcome of having two mothers at once does not count.

from itertools import combinations

class Pair():
    def __init__(self, pair):
        self.p1 = pair[0]
        self.p2 = pair[1]
        self.has_mom      = self.p1.split('_')[0] == 'm' or self.p2.split('_')[0] == 'm'
        self.has_two_moms = self.p1.split('_')[0] == 'm' and self.p2.split('_')[0] == 'm'
        self.has_daughter = self.p1.split('_')[0] == 'd' or self.p2.split('_')[0] == 'd'
        self.same_family  = self.p1.split('_')[1] == self.p2.split('_')[1]

families = [f"{member}_{number}" for member in ['m', 'f', 's', 'd'] for number in range(1, 21)]
pairs = [Pair(comb) for comb in combinations(families, r=2)]

s = 0
for pair in pairs:
    if pair.has_mom and not pair.has_two_moms:
        if not (pair.has_daughter and pair.same_family):
            s += 1
print("If two moms is not allowed:", round(s / len(pairs), 4))

s = 0
for pair in pairs:
    if pair.has_mom:
        if not (pair.has_daughter and pair.same_family):
            s += 1
print("If two moms is allowed:", round(s / len(pairs), 4))

# If two moms is not allowed: 0.3734
# If two moms is allowed: 0.4335
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    $\begingroup$ A calculation similar to van der Wolf's shows that indeed if you exclude two mothers being the winning team you get $(2*20*59)/(79*80) \approx 0.3734$ as the probability. Maybe OP can look if the questions explicitely talks about one mother instead of a mother. $\endgroup$
    – Ingix
    Apr 12 at 17:56
  • $\begingroup$ That's what I was saying, yeah. Thanks for the clarification. $\endgroup$ Apr 12 at 18:00
  • $\begingroup$ @Ingix Question says, "includes a mother but not her daughter". But answer given in book is based on one mother. $\endgroup$ Apr 13 at 9:15

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