QCAA Mathematical Methods Discrete random variables

This represents the probability of a specific sequence (e.g., Green-Green-Red) calculated as $(\frac{1}{4})^2(\frac{3}{4}) = \frac{3}{64}$. It fails to account for the $\binom{3}{2}=3$ different orders in which two green apples and one red apple can be drawn.

The probability of drawing a green apple is $p = \frac{10}{40} = \frac{1}{4}$. Using the binomial probability formula for 2 successes in 3 trials, $P(X=2) = \binom{3}{2}(\frac{1}{4})^2(\frac{3}{4})^1 = 3 \cdot \frac{1}{16} \cdot \frac{3}{4} = \frac{9}{64}$.

This value is incorrect and does not result from the standard binomial probability calculation for this scenario.

This is the probability of drawing exactly one green apple (or two red apples), calculated as $\binom{3}{1}(\frac{1}{4})^1(\frac{3}{4})^2 = \frac{27}{64}$.

The maximum value of $k$ is 4, so $n=4$. Using the binomial probability for the highest value, $P(X=4) = p^4 = \frac{16}{81}$, which implies $p=\frac{2}{3}$ and $q=1-p=\frac{1}{3}$.

This option swaps the probabilities for success and failure. If $p=\frac{1}{3}$, then $P(X=4)$ would be $(\frac{1}{3})^4 = \frac{1}{81}$, which contradicts the table value of $\frac{16}{81}$.

The random variable $X$ represents the number of successes in $n$ trials, taking values from $0$ to $n$. Since the table lists values only up to $k=4$, $n$ must be 4, not 5.

The maximum value in the distribution is $k=4$, which means $n$ cannot be 5. Additionally, the values for $p$ and $q$ would produce incorrect probabilities for the endpoints of the distribution.