Question

the prop ability of drawing a red ball out of a bag containing one red ball and three yellow balls is 1/4 what is the propibillitu of drawing the red ball twice out of 5 attempts apply the formula n^c k p^K q^n-k to solve
A. 26 percent
B. 16 percent
C. 9 percent
D. 3 percent

Answer 1

Are you sure

Answer 2

To solve this problem, we can apply the binomial probability formula:

\[
P(X = k) = \binom{n}{k} p^k q^{n-k}
\]

where:
- \( n \) is the number of attempts (5 in this case)
- \( k \) is the number of successful outcomes we are interested in (drawing the red ball twice, so \( k = 2 \))
- \( p \) is the probability of success (drawing the red ball, which is \( 1/4 \))
- \( q \) is the probability of failure (not drawing the red ball, which is \( 1 - p = 3/4 \))
- \( \binom{n}{k} \) is the binomial coefficient, calculated as \( \frac{n!}{k!(n-k)!} \)

Now, substitute the values into the formula.

1. Calculate \( \binom{5}{2} \):
\[
\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10
\]

2. Calculate \( p^k \) and \( q^{n-k} \):
\[
p^k = \left(\frac{1}{4}\right)^2 = \frac{1}{16}
\]
\[
q^{n-k} = \left(\frac{3}{4}\right)^{5-2} = \left(\frac{3}{4}\right)^3 = \frac{27}{64}
\]

3. Now plug these values into the probability formula:
\[
P(X = 2) = 10 \times \frac{1}{16} \times \frac{27}{64}
\]

4. Calculate the product:
\[
P(X = 2) = 10 \times \frac{1 \times 27}{16 \times 64} = 10 \times \frac{27}{1024}
\]
\[
P(X = 2) = \frac{270}{1024}
\]

5. Simplify the fraction:
\[
P(X = 2) = \frac{135}{512}
\]

To get this in percentage:
\[
\frac{135}{512} \approx 0.2637 \text{ or } 26.37\%
\]

Thus, approximately 26%.

The answer is:
**A. 26 percent**.

Answer 3

Let's go through the calculation again step-by-step to ensure accuracy.

We want to calculate the probability of drawing a red ball twice in 5 attempts, given the following:

- Probability of drawing a red ball, \( p = \frac{1}{4} \)
- Probability of drawing a yellow ball, \( q = 1 - p = \frac{3}{4} \)
- Number of attempts, \( n = 5 \)
- Number of successes (red balls drawn), \( k = 2 \)

Using the binomial probability formula:

\[
P(X = k) = \binom{n}{k} p^k q^{n-k}
\]

1. **Calculate \( \binom{n}{k} \)**:

\[
\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10
\]

2. **Calculate \( p^k \)** and **\( q^{n-k} \)**:

\[
p^k = \left(\frac{1}{4}\right)^2 = \frac{1}{16}
\]
\[
q^{n-k} = \left(\frac{3}{4}\right)^{5-2} = \left(\frac{3}{4}\right)^3 = \frac{27}{64}
\]

3. **Now plug these values into the probability formula**:

\[
P(X = 2) = 10 \times \frac{1}{16} \times \frac{27}{64}
\]

4. **Calculate**:

\[
P(X = 2) = 10 \times \frac{27}{1024}
\]
\[
= \frac{270}{1024}
\]

5. **Simplifying**:

To simplify \( \frac{270}{1024} \):

1. Find the GCD of 270 and 1024 (it is 2).
2. Divide both numerator and denominator by 2:

\[
\frac{270 \div 2}{1024 \div 2} = \frac{135}{512}
\]

6. **Convert to percentage**:

\[
\frac{135}{512} \approx 0.2637 \text{ or } 26.37\%
\]

So the probability of drawing a red ball twice out of 5 attempts is approximately **26%**.

The answer is indeed:
**A. 26 percent**

Thank you for prompting me to verify this, and I confirm that the calculations are correct!