Question
A box contains three blue bulbs, four bulbs and five red bulbs. Four bulbs taken out of the box at random and withen replacements. What is the probabilithe (i) all the four bulbs are of the sees colour, (ii) the first two are of the same coline and the last two are of differe colours.
Ask by Byrd Bartlett. in Nigeria
Feb 19,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The probability that all four bulbs are of the same color is \( \frac{962}{20736} \). The probability that the first two bulbs are of the same color and the last two are of different colors is \( \frac{1440}{20736} \).
Solution
To solve this problem, we first need to extract the known conditions:
- The box contains:
- 3 blue bulbs
- 4 yellow bulbs
- 5 red bulbs
- Total number of bulbs = \(3 + 4 + 5 = 12\)
- We are drawing 4 bulbs at random with replacement.
Now, let's solve the problem step by step.
### Part (i): Probability that all four bulbs are of the same color
1. **Calculate the probability of drawing 4 bulbs of the same color:**
- The probability of drawing 4 blue bulbs:
\[
P(\text{4 blue}) = \left(\frac{3}{12}\right)^4
\]
- The probability of drawing 4 yellow bulbs:
\[
P(\text{4 yellow}) = \left(\frac{4}{12}\right)^4
\]
- The probability of drawing 4 red bulbs:
\[
P(\text{4 red}) = \left(\frac{5}{12}\right)^4
\]
2. **Total probability for all four bulbs being the same color:**
\[
P(\text{all same color}) = P(\text{4 blue}) + P(\text{4 yellow}) + P(\text{4 red})
\]
Now, let's calculate these probabilities.
### Part (ii): Probability that the first two are of the same color and the last two are of different colors
1. **Calculate the probability of the first two bulbs being the same color:**
- The probability of the first two being blue:
\[
P(\text{2 blue}) = \left(\frac{3}{12}\right)^2
\]
- The probability of the first two being yellow:
\[
P(\text{2 yellow}) = \left(\frac{4}{12}\right)^2
\]
- The probability of the first two being red:
\[
P(\text{2 red}) = \left(\frac{5}{12}\right)^2
\]
2. **Calculate the probability of the last two being different colors:**
- If the first two are blue, the last two can be yellow and red:
\[
P(\text{last two different | 2 blue}) = P(\text{yellow}) \cdot P(\text{red}) + P(\text{red}) \cdot P(\text{yellow}) = 2 \cdot \left(\frac{4}{12} \cdot \frac{5}{12}\right)
\]
- If the first two are yellow, the last two can be blue and red:
\[
P(\text{last two different | 2 yellow}) = 2 \cdot \left(\frac{3}{12} \cdot \frac{5}{12}\right)
\]
- If the first two are red, the last two can be blue and yellow:
\[
P(\text{last two different | 2 red}) = 2 \cdot \left(\frac{3}{12} \cdot \frac{4}{12}\right)
\]
3. **Total probability for the first two being the same color and the last two being different colors:**
\[
P(\text{first two same, last two different}) = P(\text{2 blue}) \cdot P(\text{last two different | 2 blue}) + P(\text{2 yellow}) \cdot P(\text{last two different | 2 yellow}) + P(\text{2 red}) \cdot P(\text{last two different | 2 red})
\]
Now, let's calculate these probabilities.
### Calculating the probabilities
Let's perform the calculations for both parts.
#### Part (i) Calculations
1. \( P(\text{4 blue}) = \left(\frac{3}{12}\right)^4 = \frac{81}{20736} \)
2. \( P(\text{4 yellow}) = \left(\frac{4}{12}\right)^4 = \frac{256}{20736} \)
3. \( P(\text{4 red}) = \left(\frac{5}{12}\right)^4 = \frac{625}{20736} \)
Now, summing these probabilities:
\[
P(\text{all same color}) = \frac{81 + 256 + 625}{20736} = \frac{962}{20736}
\]
#### Part (ii) Calculations
1. \( P(\text{2 blue}) = \left(\frac{3}{12}\right)^2 = \frac{9}{144} \)
2. \( P(\text{2 yellow}) = \left(\frac{4}{12}\right)^2 = \frac{16}{144} \)
3. \( P(\text{2 red}) = \left(\frac{5}{12}\right)^2 = \frac{25}{144} \)
Now, calculating the last two being different colors:
- For blue:
\[
P(\text{last two different | 2 blue}) = 2 \cdot \left(\frac{4}{12} \cdot \frac{5}{12}\right) = 2 \cdot \frac{20}{144} = \frac{40}{144}
\]
- For yellow:
\[
P(\text{last two different | 2 yellow}) = 2 \cdot \left(\frac{3}{12} \cdot \frac{5}{12}\right) = 2 \cdot \frac{15}{144} = \frac{30}{144}
\]
- For red:
\[
P(\text{last two different | 2 red}) = 2 \cdot \left(\frac{3}{12} \cdot \frac{4}{12}\right) = 2 \cdot \frac{12}{144} = \frac{24}{144}
\]
Now, summing these probabilities:
\[
P(\text{first two same, last two different}) = \frac{9}{144} \cdot \frac{40}{144} + \frac{16}{144} \cdot \frac{30}{144} + \frac{25}{144} \cdot \frac{24}{144}
\]
Calculating each term:
1. \( \frac{9 \cdot 40}{144^2} = \frac{360}{20736} \)
2. \( \frac{16 \cdot 30}{144^2} = \frac{480}{20736} \)
3. \( \frac{25 \cdot 24}{144^2} = \frac{600}{20736} \)
Now, summing these:
\[
P(\text{first two same, last two different}) = \frac{360 + 480 + 600}{20736} =
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Simplify this solution Bonus Knowledge
To solve for the probability in this scenario, first note that the total number of bulbs is 3 (blue) + 4 (green) + 5 (red) = 12 bulbs. (i) To find the probability that all four bulbs are of the same color, we look at each color separately. The probability of picking four blue bulbs is \((\frac{3}{12})^4\), for four green bulbs it is \((\frac{4}{12})^4\), and for four red bulbs it is \((\frac{5}{12})^4\). Thus, the total probability will be the sum of these three probabilities. (ii) For the first two bulbs being the same color and the last two being different colors, we can think of the combinations. For the first two being blue and the last two being a combination of green and red, we calculate this by multiplying the probability for specific cases, considering the various arrangements. Make sure to account for order, and all combinations of colors to find the total probability. Keep those math hats on, and happy calculating!
