Ex.15.2 Q4 Probability Solution - NCERT Maths Class 10

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Question

A box contains \(12\) balls out of which \(x\) are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball? If \(6 \) more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find \(x\).

  

Text Solution

 

What is known?

A box contains  \(12 \)balls out of which are black and one ball is drawn at random from the box. If \( 6\) more black balls are put in the box, the probability of drawing a black ball is now double of what it was before.

What is unknown?

The probability that the ball drawn at random will be a black ball and the value of .

Reasoning:

First suppose the number of black balls as .Then find the total number of possible outcomes. Now, find the probability by using the formula

 Probability .

Now\( 6\) more balls are put in the box and the probability of drawing a black ball is now double of what it was before,

Now, the probability of drawing a black ball \(= 2 x\) probability of drawing black ball before.

Step:

Total Number of balls \(= 12\)

Let the number of black balls  \(= x\)

 Probability of getting black ball\[\begin{align} &= \frac{\text{ No of possible outcomes }}{\text{ Total no of outcomes }} \\&=\frac{x}{12}\end{align}\]

If \(6\) more black balls are put in the box, the probability of drawing a black ball is now double of what it was before,

Total number of balls \(=\text{ }12+6\)

Number of black balls \(=\text{ }x+\text{ }6\)

Now,

\(2 \times\) Probability of drawing black ball before \(=\) probability of drawing black ball.

\(2 \times\) Probability of drawing black ball before \[\begin{align}=\frac{\text{ No of possible outcomes }}{\text{ Total no of outcomes}}\end{align}\]

\[\begin{align} 2\left( \frac{x}{12} \right)&=\frac{x+6}{18} \\ 2x\,\,\times \,18&=12\left( x+6 \right) \\ 3x&=\,x+6 \\ 3x-x&=\,6 \\ 2x&=\,6 \\ x&=3 \end{align}\]

Number of black balls are \(3\)

  
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