Solving the Probability Puzzle: The Maximum Number of Sweets in a Bag

Suppose you have a bag containing red and green sweets, and you're picking two sweets one after the other. The probability of picking two red sweets with replacement is 10 percentage points higher than the probability of picking two red sweets without replacement. The challenge is to determine the maximum number of sweets in the bag given this condition.

Understanding the Problem

The problem consists of two interpretations:

Percentage points higher: PA is 10 percentage points higher than PB. 10 more than the original P-value: PA is 10 times PB (in its original form).

Using the second interpretation, we can solve the problem effectively.

Defining Variables

Let m represent the number of red sweets and n represent the number of green sweets. The probability of picking two red sweets with replacement is given by:

Probability with replacement: (frac{m}{m n} times frac{m}{m n} left(frac{m}{m n}right)^2)

The probability without replacement is:

Probability without replacement: (frac{m}{m n} times frac{m-1}{m n-1} frac{m(m-1)}{(m n)(m n-1)})

Formulating the Equation

According to the problem, the probability with replacement is 10 percentage points higher than the probability without replacement. In terms of the original probabilities, this can be expressed as:

(left(frac{m}{m n}right)^2 1.1 times frac{m(m-1)}{(m n)(m n-1)})

To simplify this, let s m n. Thus, we have:

(frac{m}{s} 1.1 times frac{m-1}{s-1})

Multiplying both sides by s(s-1), we get:

(m(s-1) 1.1m(s-1))

This can be simplified to:

(1.1s - m 10m / (11-m))

Which further simplifies to:

(s frac{10m}{11-m})

Given that s ≥ 3, m ≥ 2, we need to find valid integer values for m.

Solving for Maximum Number of Sweets

We test values of m from 2 to 10:

m 2: (s frac{10(2)}{11-2} frac{20}{9}) (not an integer) m 3: (s frac{10(3)}{11-3} frac{30}{8}) (not an integer) m 4: (s frac{10(4)}{11-4} frac{40}{7}) (not an integer) m 5: (s frac{10(5)}{11-5} frac{50}{6}) (not an integer) m 6: (s frac{10(6)}{11-6} frac{60}{5} 12) (an integer) m 7: (s frac{10(7)}{11-7} frac{70}{4} 35/2) (not an integer) m 8: (s frac{10(8)}{11-8} frac{80}{3}) (not an integer) m 9: (s frac{10(9)}{11-9} frac{90}{2} 45) (an integer) m 10: (s frac{10(10)}{11-10} frac{100}{1} 100) (an integer)

Among these, the maximum value for s is 100, meaning the maximum number of sweets in the bag is 100, with 10 red and 90 green sweets.

Practical Cross-Check

To verify, we perform a practical cross-check:

With 10 red and 90 green sweets:

Probability with replacement: (frac{10}{100} times frac{10}{100} frac{1}{100})

Probability without replacement: (frac{10}{100} times frac{9}{99} frac{1}{100} times frac{1}{11} frac{1}{100})

Calculating 1.1 times the probability without replacement:

1.1 × (frac{1}{100} times frac{1}{11}) (frac{1}{100})

This confirms that the probability with replacement is indeed 10 percentage points higher, satisfying the original condition.

Conclusion

The maximum number of sweets in the bag, under the given conditions, is 100. Out of these, 10 are red and 90 are green.