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Since the sum of probabilities for all possible values of a random variable must equal 1, we have: $0.2 + a + a + 0.2 + b = 1$ $2a + b = 1 - 0.4$ $2a + b = 0.6$ ...(1)
The mean (expected value) $E(X)$ is given by: $E(X) = \sum x \cdot P(x)$ $E(X) = 1(0.2) + 2(a) + 3(a) + 4(0.2) + 5(b) = 3$ $0.2 + 2a + 3a + 0.8 + 5b = 3$ $5a + 5b = 3 - 1$ $5a + 5b = 2$ $a + b = 0.4$ ...(2)
We have two equations: $2a + b = 0.6$ ...(1) $a + b = 0.4$ ...(2) Subtract equation (2) from equation (1): $(2a + b) - (a + b) = 0.6 - 0.4$ $a = 0.2$ Substitute $a = 0.2$ into equation (2): $0.2 + b = 0.4$ $b = 0.4 - 0.2$ $b = 0.2$
So, $a = 0.2$ and $b = 0.2$.
We need to find $P(X \ge 3)$, which is $P(X=3) + P(X=4) + P(X=5)$. $P(X \ge 3) = a + 0.2 + b = 0.2 + 0.2 + 0.2 = 0.6$
Final Answer: a = 0.2, b = 0.2, P(X>=3) = 0.6