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There are only red sweets and yellow sweets in a bag - Edexcel - GCSE Maths - Question 21 - 2020 - Paper 2
Question 21
There are only red sweets and yellow sweets in a bag. There are n red sweets in the bag. There are 8 yellow sweets in the bag. Sajid is going to take at random a swe... show full transcript
Worked Solution & Example Answer:There are only red sweets and yellow sweets in a bag - Edexcel - GCSE Maths - Question 21 - 2020 - Paper 2
Step 1
Show why the probability cannot be \( \frac{7}{10} \).
Answer
To determine the probability of selecting a red sweet, we start by expressing the total number of sweets in terms of ( n ) (the number of red sweets) and the known quantity of yellow sweets (8):
[ \text{Total sweets} = n + 8. ]
The probability of selecting a red sweet is thus:
[ P(Red) = \frac{n}{n + 8}. ]
According to the problem, this probability is claimed to be ( \frac{7}{10} ):
[ \frac{n}{n + 8} = \frac{7}{10}. ]
To show this cannot happen, we cross-multiply:
[ 10n = 7(n + 8) ]
Expanding leads to:
[ 10n = 7n + 56 ]
Subtracting ( 7n ) from both sides gives:
[ 3n = 56 ]
Thus, we find:
[ n = \frac{56}{3} \approx 18.67. ]
Since the number of red sweets ( n ) must be a whole number, the assumption that the probability is ( \frac{7}{10} ) is incorrect.
Step 2
Work out the number of red sweets in the bag.
Answer
Let ( n ) be the number of red sweets. After Sajid takes one red sweet, there remain ( n - 1 ) red sweets and the total number of sweets is ( (n + 8) - 1 = n + 7 ).
The probability that the second sweet Sajid takes is red will be:
[ P(Red_2 | Red_1) = \frac{n - 1}{n + 7}. ]
According to the problem, the probability that both sweets taken are red is given as ( \frac{3}{5} ):
[ P(Red_1) \times P(Red_2 | Red_1) = \frac{n}{n + 8} \times \frac{n - 1}{n + 7} = \frac{3}{5}. ]
Substituting ( P(Red_1) ) from previous calculations:
[ \frac{n}{n + 8} \times \frac{n - 1}{n + 7} = \frac{3}{5}. ]
Cross-multiplying leads to:
[ 5n(n - 1) = 3(n + 8)(n + 7). ]
Expanding both sides gives:
[ 5n^2 - 5n = 3(n^2 + 15n + 56). ]
Further expansion results in:
[ 5n^2 - 5n = 3n^2 + 45n + 168. ]
Rearranging all terms leads to:
[ 2n^2 - 50n - 168 = 0. ]
This can be simplified by dividing by 2:
[ n^2 - 25n - 84 = 0. ]
Applying the quadratic formula yields:
[ n = \frac{25 \pm \sqrt{25^2 - 4 \cdot 1 \cdot (-84)}}{2 \cdot 1} = \frac{25 \pm \sqrt{625 + 336}}{2} = \frac{25 \pm 37}{2}. ]
This results in two potential solutions:
[ n = \frac{62}{2} = 31 \quad \text{or} \quad n = \frac{-12}{2} = -6. ]
Since ( n ) must be a positive integer, we conclude that the number of red sweets in the bag is:
[ n = 31. ]