A bag has 10 red balls, 5 green balls and ‘x’ yellow balls. If the

1.	A bag has 10 red balls, 5 green balls and ‘x’ yellow balls. If the probability of drawing 2 red balls from the bag is 13/57 more than the probability of drawing 2 yellow balls then find total balls in the bag?1. 152. 253. 204. 195. 30
Answer» Correct Answer - Option 4 : 19 GIVEN: ⇒ Red balls = 10 ⇒ Green balls = 5 ⇒Yellow balls = ‘x’ ⇒ Total balls in the bag = 15 + x ASSUMPTION: Let P be the probability of drawing 2 yellow balls from the bag. CALCULATION: Case 1: when drawing 2 red balls from the bag. ⇒ ¹⁰C₂/ ^{(15 + x)}C₂ = P + 13/57 ----(1) Case 2: when drawing 2 yellow balls from the bag. ⇒^xC₂/ ^{(15 + x)}C₂ = P ----(2) Solving eq(1) and eq(2) ⇒ 10C2/ (15 + x)C2 = xC2/ (15 + x)C2 + 13/57 ⇒ (10C2 - xC2)/ (15 + x)C2 = 13/57 ⇒ ([90 - (x)(x-1)]/2) / (15 + x)(14 + x)/2 = 13/57 \(⇒\frac{(90 - {(x)(x-1))} \over 2}{\frac{(15 + x)(14 + x)}{2}} = \frac {13} {57} \) \(⇒\frac{(90 - {(x)(x-1))} }{(15 + x)(14 + x)} = \frac {13} {57} \) \(⇒{(90 - {(x)(x-1))} } \times 57= {13} \times {(15 + x)(14 + x)}\) \(⇒{(90 - {(x)(x-1))} } \times 57= {13} \times {(15 + x)(14 + x)}\) since options for yellow balls are 0, 10, 4, 15 by subtracting 15 from each option. So, we get ⇒ x = 4 ⇒ Total balls = 15 + x = 19 balls.

A bag has 10 red balls, 5 green balls and ‘x’ yellow balls. If the probability of drawing 2 red balls from the bag is 13/57 more than the probability of drawing 2 yellow balls then find total balls in the bag?1. 152. 253. 204. 195. 30

Answer» Correct Answer - Option 4 : 19

GIVEN:

⇒ Red balls = 10

⇒ Green balls = 5

⇒Yellow balls = ‘x’

⇒ Total balls in the bag = 15 + x

ASSUMPTION:

Let P be the probability of drawing 2 yellow balls from the bag.

CALCULATION:

Case 1: when drawing 2 red balls from the bag.

⇒ ¹⁰C₂/ ^{(15 + x)}C₂ = P + 13/57 ----(1)

Case 2: when drawing 2 yellow balls from the bag.

⇒^xC₂/ ^{(15 + x)}C₂ = P ----(2)

Solving eq(1) and eq(2)

⇒ 10C2/ (15 + x)C2 = xC2/ (15 + x)C2 + 13/57

⇒ (10C2 - xC2)/ (15 + x)C2 = 13/57

⇒ ([90 - (x)(x-1)]/2) / (15 + x)(14 + x)/2 = 13/57

\(⇒\frac{(90 - {(x)(x-1))} \over 2}{\frac{(15 + x)(14 + x)}{2}} = \frac {13} {57} \)

\(⇒\frac{(90 - {(x)(x-1))} }{(15 + x)(14 + x)} = \frac {13} {57} \)

\(⇒{(90 - {(x)(x-1))} } \times 57= {13} \times {(15 + x)(14 + x)}\)

since options for yellow balls are 0, 10, 4, 15 by subtracting 15 from each option.

So, we get

⇒ x = 4

⇒ Total balls = 15 + x = 19 balls.