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In a bag, there are some red, black and yellow balls. Sum of black and yellow balls is 9. Probability of selecting two red balls from that bag is 1/7 which is 250% of the probability of selecting two black balls. Find number of yellow balls in that bag if the number of black balls is even. 
  • a)
    3
  • b)
    4
  • c)
    5
  • d)
    8
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
In a bag, there are some red, black and yellow balls. Sum of black and...
Given:
Black balls + Yellow balls = 9
Probability of selecting two red balls = 1/7 = Probability of selecting two black balls 
Concept used:
A combination is the choice of r things from a set of n things without replacement and where order does not matter and is given by 
⇒  nCr = n!/{r! × (n - r)!}
Probability of an event = number of favourable outcomes/total possible outcomes
Calculation:
Let the number of red and black balls be x and y respectively
Then, probability of selecting two red balls = (xC2)/((9 + x)C2) = 1/7
⇒ x × (x - 1)/(9 + x)(8 + x) = 1/7
⇒ 7x2 - 7x = x2 + 17x + 72
⇒ 6x2 - 24x - 72 = 0 
⇒ x = 6 
∴ Number of red balls = 6
Total number of balls  = 9 + 6 = 15
Now, the probability of selecting two black balls from the bag = (yC2)/(15C2) = (1/7) × 100/250
⇒ y(y - 1)/(15 × 14) = (1/7) × (2/5)
⇒ y = 4
∴ The number of black ball = 4
∴ The number of yellow coloured ball = 15 - 6 - 4 = 5 balls.  
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Most Upvoted Answer
In a bag, there are some red, black and yellow balls. Sum of black and...
Given:
Black balls + Yellow balls = 9
Probability of selecting two red balls = 1/7 = Probability of selecting two black balls 
Concept used:
A combination is the choice of r things from a set of n things without replacement and where order does not matter and is given by 
⇒  nCr = n!/{r! × (n - r)!}
Probability of an event = number of favourable outcomes/total possible outcomes
Calculation:
Let the number of red and black balls be x and y respectively
Then, probability of selecting two red balls = (xC2)/((9 + x)C2) = 1/7
⇒ x × (x - 1)/(9 + x)(8 + x) = 1/7
⇒ 7x2 - 7x = x2 + 17x + 72
⇒ 6x2 - 24x - 72 = 0 
⇒ x = 6 
∴ Number of red balls = 6
Total number of balls  = 9 + 6 = 15
Now, the probability of selecting two black balls from the bag = (yC2)/(15C2) = (1/7) × 100/250
⇒ y(y - 1)/(15 × 14) = (1/7) × (2/5)
⇒ y = 4
∴ The number of black ball = 4
∴ The number of yellow coloured ball = 15 - 6 - 4 = 5 balls.  
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Community Answer
In a bag, there are some red, black and yellow balls. Sum of black and...
Given Information:
In the bag, there are red, black, and yellow balls. The sum of black and yellow balls is 9. The probability of selecting two red balls is 1/7, which is 250% of the probability of selecting two black balls. The number of black balls is even.

Solution:

Let's assume:
Let the number of black balls be x.
Number of yellow balls = 9 - x (as the sum of black and yellow balls is 9)

Given Probabilities:
Probability of selecting two red balls = 1/7
Probability of selecting two black balls = xC2 / (x + 9)C2
Since the probability of selecting two red balls is 250% of the probability of selecting two black balls:
1/7 = 2.5 * (xC2 / (x + 9)C2)

Solving the Equations:
1/7 = 2.5 * (x(x-1) / (x + 9)(x + 8))
Solving the above equation, we get x = 4.
Therefore, the number of yellow balls = 9 - 4 = 5.
So, the correct answer is option C) 5.
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In a bag, there are some red, black and yellow balls. Sum of black and yellow balls is 9. Probability of selecting two red balls from that bag is 1/7 which is 250% of the probability of selecting two black balls. Find number of yellow balls in that bag if the number of black balls is even.a)3b)4c)5d)8Correct answer is option 'C'. Can you explain this answer?
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