Question

A bag contains 5 red balls and some of blue balls. If the probability of a drawing a blue ball is doubled that of a ref ball . Find the number of blue balls in the bag.

Answer 1

Given that, a bag contains 5 red balls and some blue balls.

Number of red balls = 5

Let us assume that the bag contains 'x' blue balls.

Therefore, the total number of balls = x + 5

Now,

Probability = (Number of favourable outcomes)/(Total number of outcomes)

For Red balls:

Number of favourable outcomes = 5

Total number of outcomes = x + 5

P(red balls) = 5/(x + 5)

For blue balls:

Number of favourable outcomes = x

Total number of outcomes = x + 5

P(red balls) = x/(x + 5)

Also given that, the probability of a drawing a blue ball is doubled that of a ref ball.

According to question,

⇒ x/(x + 5) = 2 × [5/(x + 5)]

The denominator, throughout cancel. We left with

⇒ x = 2(5)

⇒ x = 10

Therefore,

There are 10 blue balls in the bag.

Answer 2

Given :-

• Bag has = 5 Red balls .
• And , some Blue balls .
• probability of a drawing a blue ball is doubled that of a red ball .

Solution :-

Probability :- (Favorable outcomes) / (Total number of possible outcomes.)

Let Total Blue balls = x .

→ So, Total Balls = (5 + x) .

So,

→ Probability of drawing a blue ball = x / (x + 5).

→ Probability of drawing a Red ball = 5 / (x + 5).

Now, given that, probability of a drawing a blue ball is doubled that of a red ball .

So,

→ x/(x + 5) = 2{ 5/(x + 5) }

→ x /(x + 5) = 10/(x + 5)

Cancel (x+5) from both sides denominator ,

→ x = 10 (Ans).

Hene, Number of Blue Balls are 10.