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Let $X$ be the random variable representing the number of oranges drawn in three trials. $X$ can take values 0, 1, 2, or 3.
The probability of drawing an orange in a single trial is $p = \frac{4}{10} = \frac{2}{5}$. The probability of not drawing an orange (i.e., drawing an apple) is $q = 1 - p = \frac{3}{5}$.
Since the trials are independent and the fruit is replaced each time, this is a binomial distribution with $n=3$ trials.
$P(X=0)$ is the probability of drawing no oranges (i.e., drawing apples in all three trials): $$P(X=0) = {3 \choose 0} p^0 q^3 = 1 \cdot 1 \cdot \left(\frac{3}{5}\right)^3 = \frac{27}{125}$$
$P(X=1)$ is the probability of drawing one orange: $$P(X=1) = {3 \choose 1} p^1 q^2 = 3 \cdot \left(\frac{2}{5}\right)^1 \cdot \left(\frac{3}{5}\right)^2 = 3 \cdot \frac{2}{5} \cdot \frac{9}{25} = \frac{54}{125}$$
$P(X=2)$ is the probability of drawing two oranges: $$P(X=2) = {3 \choose 2} p^2 q^1 = 3 \cdot \left(\frac{2}{5}\right)^2 \cdot \left(\frac{3}{5}\right)^1 = 3 \cdot \frac{4}{25} \cdot \frac{3}{5} = \frac{36}{125}$$
$P(X=3)$ is the probability of drawing three oranges: $$P(X=3) = {3 \choose 3} p^3 q^0 = 1 \cdot \left(\frac{2}{5}\right)^3 \cdot 1 = \frac{8}{125}$$
The probability distribution is: $X = 0: P(X=0) = \frac{27}{125}$ $X = 1: P(X=1) = \frac{54}{125}$ $X = 2: P(X=2) = \frac{36}{125}$ $X = 3: P(X=3) = \frac{8}{125}$
The expectation of the random variable $X$ is given by: $E(X) = \sum x \cdot P(X=x) = 0 \cdot \frac{27}{125} + 1 \cdot \frac{54}{125} + 2 \cdot \frac{36}{125} + 3 \cdot \frac{8}{125} = \frac{0 + 54 + 72 + 24}{125} = \frac{150}{125} = \frac{6}{5} = 1.2$
Final Answer: Probability Distribution: P(X=0) = 27/125, P(X=1) = 54/125, P(X=2) = 36/125, P(X=3) = 8/125. Expectation: 6/5 or 1.2
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