Question

Saina's duffel bag has 7 red, 23+6x green, and 12 blue shuttlecocks. If any 2 shuttlecocks are taken at random, and the probability of getting both red is 1/30, how many more green shuttlecocks are there than blue ones?​

Answer 1

ANSWER:

Given:

• A bag has 7 red, 23+6x green and 12 blue shuttlecocks.
• 2 shuttlecocks are taken out at random.
• Probability of getting both red = 1/30

To Find:

• How many more green shuttlecocks are there than blue ones.

Solution:

$:\implies\text{\sf{Total number of shuttlecocks = 7 + 23 + 6x + 12 = (42 + 6x)}}\\\\\text{\sf{As 2 shuttles are picked at random, number of outcomes-:}}\\\\:\implies\sf{^{n}C_{r} =\dfrac{(n)!}{(r)!(n-r)!}}\\\\\text{\sf{Here, n=(42+6x) and r=2. So,}}\\\\:\implies\sf{^{42+6x}C_{2}=\dfrac{(42+6x)!}{(2)!(40+6x)!}=\dfrac{(42+6x)(41+6x)(40+6x)!}{(2)(40+6x)!}}\\\\\text{\sf{As (40+6x)! gets cancelled,}}\\\\:\implies\sf{Total\:number\:of\:outcomes=^{\:42+6x}C_{2}=\dfrac{(42+6x)(41+6x)}{(2)}}$

$\text{\sf{Number of outcomes with both red shuttles:-}}\\\\:\implies\sf{^{n}C_{r} =\dfrac{(n)!}{(r)!(n-r)!}}\\\\\text{\sf{Here, n=7 and r=2. So,}}\\\\:\implies\sf{^{7}C_{2} =\dfrac{(7)!}{(2)!(7-2)!}=\dfrac{7\times6\times5!}{(2)(5)!}=\dfrac{42}{2}=21}\\\\:\implies\text{\sf{Probability of getting both red[P(R)]}}=\dfrac{No.\:of\:possible\:outcomes}{Total\:outcomes}\\\\:\implies \dfrac{1}{30} = \dfrac{\frac{21}{1}}{\frac{(42+6x)(41+6x)}{2}}\\\\:\implies\dfrac{1}{30} = \dfrac{42}{(42+6x)(41+6x)}$

$:\implies\sf{(42+6x)(41+6x) = 42*30}\\\\:\implies\sf{1722+252x+246x+36x^2 = 1260}\\\\:\implies\sf{36x^2+498x+462=0}\\\\:\implies\sf{6x^2+83x+77=0}\\\\:\implies\sf{6x^2+6x+77x+77=0}\\\\:\implies\sf{6x(x+1)+77(x+1)=0}\\\\:\implies\sf{(x+1)(6x+77)=0}\\\\:\implies\sf{x=-1\:\:or\:\:x=\dfrac{-77}{6}}\\\\:\implies\text{\sf{Number of green shuttles = 23 + 6x}}\\\\:\implies\text{\sf{Number of green shuttles = 23 + 6(-1)\:\:or\:\:23 + 6( \frac{-77}{6} )}}$

$:\implies\text{\sf{Number of green shuttles = 23 - 6\:\:or\:\:23 - 77}}\\\\:\implies\text{\sf{Number of green shuttles = 17\:\:or\:\:-54}}\\\\\text{\sf{As number of shuttles can't be negative, so:}}\\\\:\implies\text{\sf{Number of green shuttles = 17}}\\\\\text{\sf{Hence, difference of green and blue shuttles:}}\\\\:\implies\text{\sf{Green shuttles - Blue shuttles = 17 - 12 = 5}}\\\\\text{\bf{ \therefore There are 5 more green balls than blue balls.}}$

Formula Used:

   $\:\;\:\:\bullet\:\:\:\:\sf{^{n}C_{r} =\dfrac{(n)!}{(r)!(n-r)!}}$