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. 15
2. 25
3. 20
4. 19
5. 30
1. 15
2. 25
3. 20
4. 19
5. 30
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.