Solveeit Logo

Question

Question: A bag contains 3 red and 2 green balls. AN expreiment consists of drawing a ball from this bag and r...

A bag contains 3 red and 2 green balls. AN expreiment consists of drawing a ball from this bag and replacing it with 1 more ball of same colour in this bag. This experiment in done thrice. Now one more ball is drawn from the bag, Find the chance that it is red

Answer

3/5

Explanation

Solution

The problem describes a process known as Polya's Urn scheme.

Initial State:

  • Number of red balls (R0R_0) = 3
  • Number of green balls (G0G_0) = 2
  • Total number of balls (N0N_0) = R0+G0=3+2=5R_0 + G_0 = 3 + 2 = 5

The Experiment:

An experiment consists of drawing a ball, replacing it, and then adding 1 more ball of the same colour to the bag. This experiment is performed thrice.

The Question:

After these three experiments, one more ball is drawn from the bag. We need to find the probability that this ball is red.

Polya's Urn Property:

A key property of Polya's Urn scheme is that the probability of drawing a specific color ball on any given draw is the same as the initial probability of drawing that color ball.

Let P(Rk)P(R_k) be the probability of drawing a red ball on the kk-th draw.

For a Polya's Urn scheme: P(Rk)=R0N0P(R_k) = \frac{R_0}{N_0} for any k1k \ge 1.

In this problem, R0=3R_0 = 3 and N0=5N_0 = 5.

We need to find the probability of drawing a red ball on the fourth draw, i.e., P(R4)P(R_4).

According to the property of Polya's Urn:

P(R4)=P(R1)=Initial number of red ballsInitial total number of ballsP(R_4) = P(R_1) = \frac{\text{Initial number of red balls}}{\text{Initial total number of balls}}

P(R4)=35P(R_4) = \frac{3}{5}

Thus, the probability of drawing a red ball on the 4th draw is 3/5.

Explanation:

In a Polya's Urn scheme, if you start with R0R_0 red balls and G0G_0 green balls, and after drawing a ball, you replace it and add cc more balls of the same color, then the probability of drawing a red ball on any subsequent draw (kk-th draw) remains constant and is equal to the initial probability of drawing a red ball.

Initial probability of drawing a red ball = (Number of red balls) / (Total number of balls) = 3 / (3+2) = 3/5.

Since the experiment is performed thrice, and then one more ball is drawn (which is the fourth draw), the probability of this fourth ball being red is still 3/5 according to Polya's Urn property.