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

Answer 1

To find the probability of drawing a red ball twice out of 5 attempts, we can use the binomial probability formula:

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

where:
- \( n \) = total number of trials (in this case, 5 attempts),
- \( k \) = number of successful outcomes (in this case, 2 draws of the red ball),
- \( p \) = probability of success on a single trial (probability of drawing the red ball),
- \( q \) = probability of failure on a single trial (probability of not drawing the red ball).

Given:
- There is 1 red ball and 3 yellow balls, so the total number of balls is 4.
- Therefore, the probability \( p \) of drawing the red ball is \( \frac{1}{4} \) and the probability \( q \) of drawing a yellow ball is \( 1 - p = 1 - \frac{1}{4} = \frac{3}{4} \).

Now, substituting these values into the formula:

- \( n = 5 \)
- \( k = 2 \)
- \( p = \frac{1}{4} \)
- \( q = \frac{3}{4} \)

The binomial coefficient \( \binom{n}{k} \) is calculated as:

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

Now we can substitute everything into the formula:

\[
P(X = 2) = \binom{5}{2} p^2 q^{5-2}
\]

Calculating \( P(X = 2) \):

\[
P(X = 2) = 10 \left( \frac{1}{4} \right)^2 \left( \frac{3}{4} \right)^3
\]

Calculating \( \left( \frac{1}{4} \right)^2 \):

\[
\left( \frac{1}{4} \right)^2 = \frac{1}{16}
\]

Calculating \( \left( \frac{3}{4} \right)^3 \):

\[
\left( \frac{3}{4} \right)^3 = \frac{27}{64}
\]

Now, substituting these back into \( P(X = 2) \):

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

Calculating this gives:

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

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

Simplifying \( \frac{270}{1024} \):

To simplify \( \frac{270}{1024} \), we can divide the numerator and denominator by 2:

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

Thus, the probability of drawing the red ball twice out of 5 attempts is:

\[
\boxed{\frac{135}{512}}
\]