Question

A bag contains 8 red balls and some blue balls.if the probability of drawing a blue ball is thrice that of red ball , determine the number of blue balls

Answer 1

Given that:

• A bag contains 8 red balls and some blue balls.
• The probability of drawing a blue ball is thrice that of red ball.

Let us assume:

• A bag contains x blue balls.
• Total number balls in the bag = 8 + x

To Find:

• The number of blue balls in the bag.

Formula used:

• P(E) = F/T

Where,

• P = Probability
• E = Events
• F = Favourable outcomes
• T = Total outcomes

Now we have,

• Total outcomes = 8 + x
• Favourable outcomes for blue ball = x
• Favourable outcomes for red ball = 8

According to the question.

P(drawing a blue ball) = 3 × P(drawing a red ball)

ㅤ↠ㅤx/(8 + x) = 3 × 8/(8 + x)

Cancelling (8 + x) both sides.

ㅤ↠ㅤx = 3 × 8

ㅤ↠ㅤx = 24

Hence,

• There are 24 blue balls in the bag.

Answer 2

$\large \mathsf{Given :-}$

• A bag contains 8 red balls and some blue balls.

• The probability of drawing a blue ball is thrice that of red ball.

$\large \mathsf{Let's \: assume :-}$

• A bag contains x blue balls.

• Total number balls in the bag = 8 + x

$\large \mathsf{To \: Find :-}$

• The number of blue balls in the bag.

$\large \mathsf{Formula \: used :-}$

• ☆ P(E) = Favourable outcomes/ Total outcomes ☆

$\large \mathsf{Now \: we \: have,}$

• Total outcomes = 8 + x

• Favourable outcomes for blue ball = x

• Favourable outcomes for red ball = 8

★ According to the question ★

$\large \pink{ \large \underline{ \large \mathsf{{ \large \boxed{ \mathsf{P(drawing \: a \: blue \: ball) = 3 × P(drawing \: a \: red \: ball)}}}}}}$

$\longrightarrow \huge \mathsf{ \frac{x}{(8 + x)} = \frac{3 \times 8}{(8 + x)} }$

• Cancelling (8 + x) both sides.

ㅤ

$\longrightarrow \: \huge \mathsf{x = 3 × 8}$

$\longrightarrow \: \huge \boxed{ \mathsf \orange{x = 24}}$

Hence,

• There are 24 blue balls in the bag.