In a bag, there are total 15 balls of red and blue color. Find the number of blue balls, if probability of getting both balls of same color when two balls are picked randomly is $\frac{17}{35}$. (Blue balls > Red balls)

Correct Option: 9

Step-by-Step Explanation:

Let the total number of blue balls in the bag be b.
Since the total number of balls is 15, the number of red balls is (15 - b).

We are given that the number of blue balls is greater than the number of red balls (Blue balls > Red balls), which means:

$b>15-b$

$2b>15\Rightarrow b>7.5$

Thus, b must be an integer greater than 7.5 (i.e., b ≥ 8).

Total ways to choose 2 balls out of 15:

The total number of ways to pick any 2 balls from 15 balls is given by the combination formula ¹⁵C₂:

${}^{15}{\text{C}}_2=\frac{15\times 14}{2}=105$

Favorable ways to pick 2 balls of the same color:

Two balls picked will have the same color if both are blue OR both are red.

• Ways to pick 2 blue balls from b blue balls = ^bC₂
• Ways to pick 2 red balls from (15 - b) red balls = ^{(15 - b)}C₂

Total favorable outcomes = ^bC₂ + ^{(15 - b)}C₂

Setting up the probability equation:

The probability of picking two balls of the same color is given as ¹⁷/₃₅.

$\frac{{}^b{\text{C}}_2+{}^{15-b}{\text{C}}_2}{105}=\frac{17}{35}$

Multiply both sides by 105 to solve for the total favorable outcomes:

${}^b{\text{C}}_2+{}^{15-b}{\text{C}}_2=\frac{17}{35}\times 105$

${}^b{\text{C}}_2+{}^{15-b}{\text{C}}_2=17\times 3=51$

Expanding the combinations:

$\frac{b(b-1)}{2}+\frac{(15-b)(14-b)}{2}=51$

Multiply the entire equation by 2:

$b(b-1)+(15-b)(14-b)=102$

Expand both terms:

$(b^2-b)+(210-29b+b^2)=102$

Combine like terms:

$2b^2-30b+210=102$

$2b^2-30b+108=0$

Divide the quadratic equation by 2:

$b^2-15b+54=0$

Factorize the equation:

$(b-9)(b-6)=0$

So, b = 9 or b = 6.

Since we are given that Blue balls > Red balls (b > 7.5), we reject b = 6 and select b = 9.

Therefore, the total number of blue balls in the bag is 9.