Question

An urn contains 4 red and 7 black balls. Two balls are drawn at random with replacement. Find the probability of getting
(i) 2 red balls,
(ii) 2 black balls,
(iii) one red and one black ball.

Answer 1

Hint: Find the total balls after each drawing. Then as you know number balls of red and number balls of black. Find the probability at each drawing and multiply the 2 drawings with the probabilities as asked in the question: red or black. Solve the 3 cases in the same manner as described above. The main point in question is that replacement is followed.

Complete step-by-step answer:

Given in question “with replacement” that means the balls are drawn and then again kept inside the urn as it is.

So, after each drawing the total number of balls will remain 11 always.

So, after each drawing the total number of red balls will remain 4 always.

So, after each drawing the total number of black balls will remain 7 always.

Let probability of drawing red be P(R) and probability of drawing black be P(B).

$\text{Probability = }\dfrac{\text{Favourable cases}}{\text{Total cases}}$

By using the above formula, we get probabilities as:

$\begin{align}

  & P\left( R \right)=\dfrac{4}{11} \\

 & P\left( B \right)=\dfrac{7}{11} \\

\end{align}$

Case 1: Probability of drawing 2 red balls, will be

Let us assume that the probability of this will be as P(2R).

P(2R) = P(R) x P(R)

This is because the probability must be Red, Red

$P\left( 2R \right)=\dfrac{4}{11}\times \dfrac{4}{11}=\dfrac{16}{121}$

Case 2: Probability of drawing 2 black balls will be:

Let us assume that the probability of this will be as P(2B).

P(2B) = P(B) x P(B)

This is because the probability must be black, black.

$P\left( 2B \right)=\dfrac{7}{11}\times \dfrac{7}{11}=\dfrac{49}{121}$

Case 3: Probability of drawing 1 red, 1 black ball will be:

Let us assume that the probability of this will be P(RB).

P(RB) = P(R) x P(B) + P(B) x P(R)

This is because the probability can be R, B and B, R.

$P\left( RB \right)=\dfrac{7}{11}\times \dfrac{4}{11}\times 2=\dfrac{56}{121}$

Therefore, all probabilities which are asked for are all calculated.

Note: In the last case there are 2 possibilities, red coming first or black coming first. So, don’t forget to multiply with 2 at last.