Let A, B and C be three mutually exclusive and exhaustive events associated with a random experiment. If P (B) = 1.5 P (A) and P (C) = 0.5 P (B), then P (A) is equal to
4/13
Concept:
- A, B, and C are mutually exclusive and collectively exhaustive events related to a random experiment.
- P (A ∪ B ∪ C) = P (A) + P (B) + P (C)
- If A ∪ B ∪ C = S = sample space ⇔ P (A ∪ B ∪ C) = 1
Calculation:
Given:
P (B) = 1.5 P (A) and P (C) = 0.5 P (B) and
A, B, and C are mutually exclusive and collectively exhaustive events related to a random experiment.
⇒ P (A ∪ B ∪ C) = 1
⇒ P (A) + P (B) + P (C) = 1
⇒ P (A) + 1.5 P (A) + 0.5 P (B) = 1
⇒ 2.5 P (A) + 0.5 × 1.5 P (A) = 1
⇒ 2.5 P (A) + 0.75 P (A) = 1
⇒ 3.25 P (A) = 1
⇒ P (A) = 1/3.25 = 4/13
∴ Option 2 is correct.
Important Questions from Probability of Random Experiments
A, B, C and D are mutually exclusive and exhaustive events.
If 2P(A) = 3P(B) = 4P(C) = 5P(D), then what is 77P(A) equal to ?
A fair coin is tossed 6 times. What is the probability of getting a result in the 6t h toss which is different from those obtained in the first five tosses ?
Two cards are drawn successively without replacement from a well-shuffled pack of 52 cards. The probability of drawing two aces is
A biased coin with the probability of getting head equal to \(\frac{1}{4}\) is tossed five times. What is the probability of getting tail in all the first four tosses followed by head ?
Three dice are thrown. What is the probability that each face shows only multiples of 3 ?
