Question

A bag contams 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, find the number of blue balls in the bag.​

Answer 1

Answer:

let blue balls=x

red balls =5

Total balls =5+x

p(blue ball)=x/5+x

P(red ball)=5/5+x

A.T.Q

 x/5+x=2×5/5+x

x/5+x=10/5+x

x=10

if you satisfy with my answer

breanielist..............

Step-by-step explanation:

Answer 2

Let the -

• number of blue balls be x.

A bad contains 5 red balls and some blue balls.

Probability of drawing blue ball is double that of red ball.

So,

⇒ Red balls = 5

⇒ Total number of balls = x + 5

$\sf{ Probability \:of \:red \:balls\: =\:{\frac{Number\:red\:balls}{Total\:number \:of\:balls}}}$

→ $\sf{\frac{5}{x\:+\:5}}$

$\sf{ Probability \:of \:blue \:balls\: =\:{\frac{Number\:blue\:balls}{Total\:number \:of\:balls}}}$

→ $\sf{\frac{x}{x\:+\:5}}$

According to question,

→ $\sf{2(\frac{5}{x\:+\:5})\:=\:\frac{x}{x\:+\:5}}$

→ $\sf{\frac{10}{x\:+\:5}\:=\:\frac{x}{x\:+\:5}}$

→ $\sf{10(x\:+\:5)\:=\:x(x\:+\:5)}$

→ $\sf{10x\:+\:50\:=\:x^2\:+\:5x}$

→ $\sf{x^2\:-\:5x\:-\:50\:=\:0}$

→ $\sf{x^2\:-\:10x\:+\:5x\:-\:50\:=\:0}$

→ $\sf{x(x\:-\:10)\:+5(x\:-\:10)\:=\:0}$

→ $\sf{(x\:+\:5)\:(x\:-\:10)\:=\:0}$

→ $\sf{x\:=\:-5(neglected),\:10}$

So,

Blue balls = 10

Total number of balls = 5 + 10 = 15

•°• Number of blue balls = 10