Question

A box contains 4 Red and 3 White balls. Every time one ball is drawn randomly and is placed back along with two more balls of opposite colour. What is the probability that among first 3 trials in first two one get red colour ball and in 3 rd he get white ball. (a) \((8 / 27)\) (b) \((16 / 99)\) (c) \((16 / 231)\) (d) none

Short answer

The probability of drawing a red ball in the first two trials and a white ball in the third trial is \(\frac{3}{28}\), which doesn't match any of the given options. Hence, the correct answer is (d) None.

Step-by-step solution

Determine the initial probabilities for the first trial

In this step, calculate the probability of drawing a red ball and a white ball in the first trial. There are a total of 4 red balls and 3 white balls, so there are 4+3=7 balls in the box. The probability of drawing a red ball in the first trial: \(P(R_1) = \frac{4}{7}\) The probability of drawing a white ball in the first trial: \(P(W_1) = \frac{3}{7}\)

Determine the probabilities for the second trial

In this step, calculate the probability of drawing a red ball and a white ball in the second trial, given that there are different probabilities resulting from the first trial. If the first drawn ball is red, then there will be 3 red balls and \(3+2=5\) white balls in the box. If the first drawn ball is white, then there will be \(4+2=6\) red balls and 3 white balls in the box. If the first drawn ball is red, the probability of drawing a red ball in the second trial: \(P(R_2|R_1) = \frac{3}{8}\) If the first drawn ball is red, the probability of drawing a white ball in the second trial: \(P(W_2|R_1) = \frac{5}{8}\) If the first drawn ball is white, the probability of drawing a red ball in the second trial: \(P(R_2|W_1) = \frac{6}{9}\) If the first drawn ball is white, the probability of drawing a white ball in the second trial: \(P(W_2|W_1) = \frac{3}{9}\)

Calculate the probability

We want to calculate the probability of drawing a red ball in the first two trials and a white ball in the third trial: \(P(R_1R_2W_3) = P(R_1)\times P(R_2|R_1)\times P(W_3|R_1R_2)\) From step 1 and step 2, we substitute the calculated values for the probabilities: \(P(R_1R_2W_3) = P(R_1)\times P(R_2|R_1)\times P(W_3|R_1R_2) = \frac{4}{7}\times \frac{3}{8}\times \frac{5}{10}\) Now, we can calculate the probability: \(P(R_1R_2W_3) = \frac{60}{560}\) To simplify, we reduce the fraction: \(P(R_1R_2W_3) = \frac{6}{56} = \frac{3}{28}\) However, none of the given options match our calculated probability. Therefore, the correct option is: (d) None

Key concepts

conditional probability

Conditional probability is a fundamental concept in probability theory that helps us analyze the probability of an event, given that another event has already occurred. This allows us to refine our probability estimates based on additional information.
For instance, in our problem, the probability of drawing a ball changes depending on the color of the previously drawn ball.

• If the first ball drawn is red, the conditions for the second draw change because now there are fewer red balls and more white balls in the box.
• This adjustment of probabilities is what conditional probability addresses, where \( P(A|B) \), the probability of event A given event B, is calculated as the probability of A and B both happening divided by the probability of B.

This is crucial when calculating the final probability of drawing two red followed by one white ball in sequence because each step depends on the outcome of the previous draw.

combinatorial analysis

Combinatorial analysis involves counting, arranging, and analyzing possible outcomes in a structured way. It is deeply connected to probability, as probability often arises from the analysis of various combinations of outcomes.
In our exercise, combinatorial thinking helps us determine all possible sequences of drawing balls, but it is simplified by considering only relevant sequences due to the stipulated conditions and replacements.

• Initially, we calculate the outcomes based on the existing conditions.
• We then adjust the number of possible outcomes whenever a ball is added back into the box with two of another color.

Combinatorial analysis in this problem helps deduce the potential sequences leading to the final desired outcome, simplifying the calculation of step-by-step probabilities rather than calculating an overwhelming number of all possible arrangements.

independent events

Independent events in probability have no influence on one another. Knowing the outcome of one does not change the probability of another. Contrastingly, the events in our problem are not independent.
In this exercise, because each drawn ball affects the subsequent draw's setting (by changing the number of each color in the box), each draw is affected by the previous one. This dependency is why conditional probability (discussed earlier) is necessary.

• The dependency is evident, as the number of possible red or white balls changes the subsequent draw's probabilities.
• If the events were independent, such conditions wouldn't alter future probabilities, simplifying the calculations significantly.

Thus, while many probability problems may assume or involve independence for simplification, our exercise must effectively manage these dependencies to calculate the correct probabilities.