To find the probability of drawing a red ball twice out of five attempts from a bag containing one red ball and three yellow balls, we first need to determine the probability of drawing a red ball in a single attempt.
Given that there is 1 red ball and 3 yellow balls, the total number of balls is:
\[ 1 + 3 = 4 \]
Thus, the probability \( P(R) \) of drawing a red ball in one attempt is:
\[ P(R) = \frac{\text{Number of red balls}}{\text{Total number of balls}} = \frac{1}{4} \]
The probability of drawing a yellow ball \( P(Y) \) is:
\[ P(Y) = \frac{3}{4} \]
Next, we want to find the probability of drawing a red ball exactly twice in five attempts. This situation can be described using the binomial probability formula:
\[ P(X = k) = \binom{n}{k} (p)^k (1 - p)^{n - k} \]
where:
- \( n \) is the number of trials (in this case, \( n = 5 \)),
- \( k \) is the number of successful trials (in this case, \( k = 2 \)),
- \( p \) is the probability of success on a single trial (in this case, \( p = \frac{1}{4} \)),
- \( \binom{n}{k} \) is the binomial coefficient.
First, we calculate the binomial coefficient \( \binom{5}{2} \):
\[ \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \]
Now we can apply the binomial formula:
\[ P(X = 2) = \binom{5}{2} \left(\frac{1}{4}\right)^2 \left(\frac{3}{4}\right)^{5-2} \]
Calculating \( \left(\frac{1}{4}\right)^2 \):
\[ \left(\frac{1}{4}\right)^2 = \frac{1}{16} \]
Calculating \( \left(\frac{3}{4}\right)^3 \):
\[ \left(\frac{3}{4}\right)^3 = \frac{27}{64} \]
Putting it all together:
\[ P(X = 2) = 10 \times \frac{1}{16} \times \frac{27}{64} \]
Calculating \( 10 \times \frac{1}{16} \):
\[ 10 \times \frac{1}{16} = \frac{10}{16} = \frac{5}{8} \]
Now, we multiply this by \( \frac{27}{64} \):
\[ P(X = 2) = \frac{5}{8} \times \frac{27}{64} = \frac{135}{512} \]
Therefore, the probability of drawing the red ball exactly twice in five attempts is:
\[ \frac{135}{512} \]
So, the final answer is:
\[ \text{Probability} = \frac{135}{512} \]
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