Step 1: Write the given equation:
The total probability is given as: \[ P(0) + P(1) + P(2) + P(3) + P(4) = 1 \] Substituting the known values: \[ 0.1 + k + 2k + k + 0.1 = 1 \] where \( P(1) = k \), \( P(2) = 2k \), and \( P(3) = k \).
Step 2: Simplify the equation:
Combine the terms: \[ 0.2 + 4k = 1 \] Subtract \( 0.2 \) from both sides: \[ 4k = 0.8 \] Divide by \( 4 \) to find \( k \): \[ k = 0.2 = \frac{1}{5} \]
Step 3: Find \( P(2) \): Given \( P(2) = 2k \), substitute the value of \( k \): \[ P(2) = 2 \times \frac{1}{5} = \frac{2}{5} \]
Conclusion: The value of \( P(2) \) is \( \mathbf{\frac{2}{5}} \).
Determine whether each of the following relations are reflexive, symmetric, and transitive.
Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.
ASSERTION-REASON BASED QUESTIONS:-
Questions number 19 and 20 are Assertion and Reason based questions.
Two statements are given, one labelled Assertion (A) and the other labelled Reason (R).
Select the correct answer from the codes (A), (B), (C) and (D) as given below.