Correct answer: \(\frac{3}{5}\)
Explanation:
The total number of balls in the bag is: \[ 4 \, \text{(black balls)} + 6 \, \text{(red balls)} = 10 \, \text{balls} \] The number of red balls is 6. Therefore, the probability of drawing a red ball is: \[ P(\text{red ball}) = \frac{\text{number of red balls}}{\text{total number of balls}} = \frac{6}{10} = \frac{3}{5} \]
Hence, the probability of drawing a red ball is \(\frac{3}{5}\).
If A is any event associated with sample space and if E1, E2, E3 are mutually exclusive and exhaustive events. Then which of the following are true?
(A) \(P(A) = P(E_1)P(E_1|A) + P(E_2)P(E_2|A) + P(E_3)P(E_3|A)\)
(B) \(P(A) = P(A|E_1)P(E_1) + P(A|E_2)P(E_2) + P(A|E_3)P(E_3)\)
(C) \(P(E_i|A) = \frac{P(A|E_i)P(E_i)}{\sum_{j=1}^{3} P(A|E_j)P(E_j)}, \; i=1,2,3\)
(D) \(P(A|E_i) = \frac{P(E_i|A)P(E_i)}{\sum_{j=1}^{3} P(E_i|A)P(E_j)}, \; i=1,2,3\)
Choose the correct answer from the options given below: