A box contains three blue bulbs, fourg goth bulbs and five red bulbs. Four hult ity taken out of the box at random and willones replacements. What is the probability by (i) all the four bulbs are of the same colour; (ii) the first two are of the same colos and the last two are of differem colours.

To solve this problem, we first need to extract the known conditions: - The box contains: - 3 blue bulbs - 4 green bulbs - 5 red bulbs - Total bulbs = \(3 + 4 + 5 = 12\) - We are drawing 4 bulbs with replacement. Now, let's solve the two parts of the problem step by step. ### Part (i): Probability that all four bulbs are of the same color To find the probability that all four bulbs drawn are of the same color, we need to consider the three colors separately. 1. **Probability of drawing 4 blue bulbs:** \[ P(\text{4 blue}) = \left(\frac{3}{12}\right)^4 \] 2. **Probability of drawing 4 green bulbs:** \[ P(\text{4 green}) = \left(\frac{4}{12}\right)^4 \] 3. **Probability of drawing 4 red bulbs:** \[ P(\text{4 red}) = \left(\frac{5}{12}\right)^4 \] Now, we can sum these probabilities to find the total probability of drawing 4 bulbs of the same color: \[ P(\text{all same color}) = P(\text{4 blue}) + P(\text{4 green}) + P(\text{4 red}) \] Let's calculate this step by step. ### Part (ii): Probability that the first two are of the same color and the last two are of different colors To find this probability, we can break it down into cases based on the color of the first two bulbs. 1. **Case 1: First two are blue** - Probability of first two being blue: \(\left(\frac{3}{12}\right)^2\) - Probability of the third bulb being either green or red: \(\frac{4 + 5}{12} = \frac{9}{12}\) - Probability of the fourth bulb being the other color (not blue): \(\frac{4 + 5 - 1}{12} = \frac{8}{12}\) So, the probability for this case is: \[ P(\text{first two blue}) = \left(\frac{3}{12}\right)^2 \cdot \frac{9}{12} \cdot \frac{8}{12} \] 2. **Case 2: First two are green** - Probability of first two being green: \(\left(\frac{4}{12}\right)^2\) - Probability of the third bulb being either blue or red: \(\frac{3 + 5}{12} = \frac{8}{12}\) - Probability of the fourth bulb being the other color (not green): \(\frac{3 + 5 - 1}{12} = \frac{7}{12}\) So, the probability for this case is: \[ P(\text{first two green}) = \left(\frac{4}{12}\right)^2 \cdot \frac{8}{12} \cdot \frac{7}{12} \] 3. **Case 3: First two are red** - Probability of first two being red: \(\left(\frac{5}{12}\right)^2\) - Probability of the third bulb being either blue or green: \(\frac{3 + 4}{12} = \frac{7}{12}\) - Probability of the fourth bulb being the other color (not red): \(\frac{3 + 4 - 1}{12} = \frac{6}{12}\) So, the probability for this case is: \[ P(\text{first two red}) = \left(\frac{5}{12}\right)^2 \cdot \frac{7}{12} \cdot \frac{6}{12} \] Now, we can sum these probabilities to find the total probability for part (ii): \[ P(\text{first two same, last two different}) = P(\text{first two blue}) + P(\text{first two green}) + P(\text{first two red}) \] Let's calculate these probabilities. ### Calculations Now, I will perform the calculations for both parts. #### Part (i) Calculation 1. \(P(\text{4 blue}) = \left(\frac{3}{12}\right)^4\) 2. \(P(\text{4 green}) = \left(\frac{4}{12}\right)^4\) 3. \(P(\text{4 red}) = \left(\frac{5}{12}\right)^4\) #### Part (ii) Calculation 1. \(P(\text{first two blue}) = \left(\frac{3}{12}\right)^2 \cdot \frac{9}{12} \cdot \frac{8}{12}\) 2. \(P(\text{first two green}) = \left(\frac{4}{12}\right)^2 \cdot \frac{8}{12} \cdot \frac{7}{12}\) 3. \(P(\text{first two red}) = \left(\frac{5}{12}\right)^2 \cdot \frac{7}{12} \cdot \frac{6}{12}\) Let's perform these calculations. Calculate the value by following steps: - step0: Calculate: \(\left(\frac{3}{12}\right)^{4}+\left(\frac{4}{12}\right)^{4}+\left(\frac{5}{12}\right)^{4}\) - step1: Reduce the fraction: \(\left(\frac{1}{4}\right)^{4}+\left(\frac{4}{12}\right)^{4}+\left(\frac{5}{12}\right)^{4}\) - step2: Reduce the fraction: \(\left(\frac{1}{4}\right)^{4}+\left(\frac{1}{3}\right)^{4}+\left(\frac{5}{12}\right)^{4}\) - step3: Evaluate the power: \(\frac{1}{256}+\left(\frac{1}{3}\right)^{4}+\left(\frac{5}{12}\right)^{4}\) - step4: Evaluate the power: \(\frac{1}{256}+\frac{1}{81}+\left(\frac{5}{12}\right)^{4}\) - step5: Evaluate the power: \(\frac{1}{256}+\frac{1}{81}+\frac{625}{20736}\) - step6: Reduce fractions to a common denominator: \(\frac{81}{256\times 81}+\frac{256}{81\times 256}+\frac{625}{20736}\) - step7: Multiply the numbers: \(\frac{81}{20736}+\frac{256}{81\times 256}+\frac{625}{20736}\) - step8: Multiply the numbers: \(\frac{81}{20736}+\frac{256}{20736}+\frac{625}{20736}\) - step9: Transform the expression: \(\frac{81+256+625}{20736}\) - step10: Add the numbers: \(\frac{962}{20736}\) - step11: Reduce the fraction: \(\frac{481}{10368}\) Calculate or simplify the expression \( (3/12)^2 * (9/12) * (8/12) + (4/12)^2 * (8/12) * (7/12) + (5/12)^2 * (7/12) * (6/12) \). Calculate the value by following steps: - step0: Calculate: \(\left(\frac{3}{12}\right)^{2}\left(\frac{9}{12}\right)\left(\frac{8}{12}\right)+\left(\frac{4}{12}\right)^{2}\left(\frac{8}{12}\right)\times \frac{7}{12}+\left(\frac{5}{12}\right)^{2}\times \frac{7}{12}\left(\frac{6}{12}\right)\) - step1: Reduce the fraction: \(\left(\frac{1}{4}\right)^{2}\left(\frac{9}{12}\right)\left(\frac{8}{12}\right)+\left(\frac{4}{12}\right)^{2}\left(\frac{8}{12}\right)\times \frac{7}{12}+\left(\frac{5}{12}\right)^{2}\times \frac{7}{12}\left(\frac{6}{12}\right)\) - step2: Reduce the fraction: \(\left(\frac{1}{4}\right)^{2}\left(\frac{9}{12}\right)\left(\frac{8}{12}\right)+\left(\frac{1}{3}\right)^{2}\left(\frac{8}{12}\right)\times \frac{7}{12}+\left(\frac{5}{12}\right)^{2}\times \frac{7}{12}\left(\frac{6}{12}\right)\) - step3: Reduce the fraction: \(\left(\frac{1}{4}\right)^{2}\times \frac{3}{4}\left(\frac{8}{12}\right)+\left(\frac{1}{3}\right)^{2}\left(\frac{8}{12}\right)\times \frac{7}{12}+\left(\frac{5}{12}\right)^{2}\times \frac{7}{12}\left(\frac{6}{12}\right)\) - step4: Reduce the fraction: \(\left(\frac{1}{4}\right)^{2}\times \frac{3}{4}\times \frac{2}{3}+\left(\frac{1}{3}\right)^{2}\left(\frac{8}{12}\right)\times \frac{7}{12}+\left(\frac{5}{12}\right)^{2}\times \frac{7}{12}\left(\frac{6}{12}\right)\) - step5: Reduce the fraction: \(\left(\frac{1}{4}\right)^{2}\times \frac{3}{4}\times \frac{2}{3}+\left(\frac{1}{3}\right)^{2}\times \frac{2}{3}\times \frac{7}{12}+\left(\frac{5}{12}\right)^{2}\times \frac{7}{12}\left(\frac{6}{12}\right)\) - step6: Reduce the fraction: \(\left(\frac{1}{4}\right)^{2}\times \frac{3}{4}\times \frac{2}{3}+\left(\frac{1}{3}\right)^{2}\times \frac{2}{3}\times \frac{7}{12}+\left(\frac{5}{12}\right)^{2}\times \frac{7}{12}\times \frac{1}{2}\) - step7: Multiply: \(\frac{1}{32}+\left(\frac{1}{3}\right)^{2}\times \frac{2}{3}\times \frac{7}{12}+\left(\frac{5}{12}\right)^{2}\times \frac{7}{12}\times \frac{1}{2}\) - step8: Multiply: \(\frac{1}{32}+\frac{7}{162}+\left(\frac{5}{12}\right)^{2}\times \frac{7}{12}\times \frac{1}{2}\) - step9: Multiply: \(\frac{1}{32}+\frac{7}{162}+\frac{175}{3456}\) - step10: Reduce fractions to a common denominator: \(\frac{81\times 4}{32\times 81\times 4}+\frac{7\times 16\times 4}{162\times 16\times 4}+\frac{175\times 3}{3456\times 3}\) - step11: Multiply the terms: \(\frac{81\times 4}{10368}+\frac{7\times 16\times 4}{162\times 16\times 4}+\frac{175\times 3}{3456\times 3}\) - step12: Multiply the terms: \(\frac{81\times 4}{10368}+\frac{7\times 16\times 4}{10368}+\frac{175\times 3}{3456\times 3}\) - step13: Multiply the numbers: \(\frac{81\times 4}{10368}+\frac{7\times 16\times 4}{10368}+\frac{175\times 3}{10368}\) - step14: Transform the expression: \(\frac{81\times 4+7\times 16\times 4+175\times 3}{10368}\) - step15: Multiply the numbers: \(\frac{324+7\times 16\times 4+175\times 3}{10368}\) - step16: Multiply the terms: \(\frac{324+448+175\times 3}{10368}\) - step17: Multiply the numbers: \(\frac{324+448+525}{10368}\) - step18: Add the numbers: \(\frac{1297}{10368}\) ### Results 1. **Part (i): Probability that all four bulbs are of the same color** - The probability is approximately \(0.046393\) or in fractional form, \(\frac{481}{10368}\). 2. **Part (ii): Probability that the first two are of the same color and the last two are of different colors** - The probability is approximately \(0.125096\) or in fractional form, \(\frac{1297}{10368}\). ### Summary - **Probability of all four bulbs being the same color:** \(\approx 0.046393\) or \(\frac{481}{10368}\) - **Probability of the first two being the same color and the last two being different colors:** \(\approx 0.125096\) or \(\frac{1297}{10368}\)