When working with probability distributions, always ensure that the sum of all probabilities equals 1. When solving for unknown probabilities, use the given conditions and simplify the equations step-by-step. It's important to confirm the validity of each probability and use correct substitutions to avoid errors. Additionally, for cumulative probabilities, remember to add up the relevant probabilities as required by the problem.
The sum of all probabilities must equal 1:
\[ P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 1. \]
Substitute the given probabilities:
\[ 0.1 + c(1) + c(2) + c(2) + c(1) = 1. \]
Simplify:
\[ 0.1 + 6c = 1 \Rightarrow 6c = 0.9 \Rightarrow c = 0.15. \]
(A) \( c = 0.15 \). Match: (A) → (IV).
(B) \( P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \):
\[ P(X \leq 2) = 0.1 + c(1) + c(2) = 0.1 + 0.15 + 0.3 = 0.55. \]
Match: (B) → (III).
(C) \( P(X = 2) = c(2) = 0.3. \) Match: (C) → (II).
(D) \( P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4) \):
\[ P(X \geq 2) = c(2) + c(2) + c(1) = 0.3 + 0.3 + 0.15 = 0.75. \]
Match: (D) → (I).
(A) - (IV), (B) - (III), (C) - (II), (D) - (I).
The sum of all probabilities must equal 1:
\[ P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 1. \]
Substitute the given probabilities:
\[ 0.1 + c(1) + c(2) + c(2) + c(1) = 1. \]
Simplify:
\[ 0.1 + 6c = 1 \Rightarrow 6c = 0.9 \Rightarrow c = 0.15. \]
Conclusion:
(A) \( c = 0.15 \). Match: (A) → (IV).
For (B) \( P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \):
\[ P(X \leq 2) = 0.1 + c(1) + c(2) = 0.1 + 0.15 + 0.3 = 0.55. \] Match: (B) → (III).
For (C) \( P(X = 2) = c(2) = 0.3. Match: (C) → (II).
For (D) \( P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4) \):
\[ P(X \geq 2) = c(2) + c(2) + c(1) = 0.3 + 0.3 + 0.15 = 0.75. \] Match: (D) → (I).
Final Matching: (A) → (IV), (B) → (III), (C) → (II), (D) → (I)
Given three identical bags each containing 10 balls, whose colours are as follows:
Bag I | 3 Red | 2 Blue | 5 Green |
Bag II | 4 Red | 3 Blue | 3 Green |
Bag III | 5 Red | 1 Blue | 4 Green |
A person chooses a bag at random and takes out a ball. If the ball is Red, the probability that it is from Bag I is $ p $ and if the ball is Green, the probability that it is from Bag III is $ q $, then the value of $ \frac{1}{p} + \frac{1}{q} $ is: