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Independent and Dependent Events

Let Xϵ 0, 1 a...

Question

$Let X ϵ (0, 1) and Y ϵ (0, 1)$ be two independent binary random variables. If P(X = 0) = p and P(Y = 0) = q, then $P (X + Y \geq 1)$ is equal to

pq + (1 - p) (1 - q)

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p(1 - q)

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1 - pq

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Solution

The correct option is D 1 - pq
X and Y can take values according to the following cases:
Either (X = 0, Y = 0) or (X = 0, Y = 1) or (X = 1, Y = 0) or (X = 1, Y = 1)

As they are binary Random variables so we can not take fractional values for X & Y

$Now, (X + Y) \geq 1 is possible only if X \neq 0 and Y \neq 0$

$P (X + Y \geq 1) = 1 - P [(X = 0) \cap (Y = 0)]$
$= 1 - P (X = 0) . P (Y = 0)$
$= (1 - p q)$

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Let Xϵ (0, 1) and Yϵ (0, 1) be two independent binary random variables. If P(X = 0) = p and P(Y = 0) = q, then P(X+Y≥1) is equal to

$Let X ϵ (0, 1) and Y ϵ (0, 1)$ be two independent binary random variables. If P(X = 0) = p and P(Y = 0) = q, then $P (X + Y \geq 1)$ is equal to