AP EAMCET · Maths · Probability
If \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) are three independent events of a random experiment such that \(\mathrm{P}\left(\mathrm{A} \cap \mathrm{B}^{\mathrm{c}} \cap \mathrm{C}^{\mathrm{c}}\right)=\frac{1}{4}, \mathrm{P}\left(\mathrm{A}^{\mathrm{c}} \cap \mathrm{B} \cap \mathrm{C}^{\mathrm{c}}\right)=\frac{1}{8}\) and \(\mathrm{P}\left(\mathrm{A}^{\mathrm{c}} \cap \mathrm{B}^{\mathrm{c}} \cap \mathrm{C}^{\mathrm{c}}\right)=\frac{1}{4}\), then \(\mathrm{P}(\mathrm{A}), \mathrm{P}(\mathrm{B})\) and \(\mathrm{P}(\mathrm{C})\) are respectively
- A \(\frac{1}{2}, \frac{1}{4}, \frac{1}{5}\)
- B \(1, \frac{1}{2}, \frac{1}{3}\)
- C \(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}\)
- D \(\frac{1}{3}, \frac{1}{4}, \frac{1}{5}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}\)
Step-by-step Solution
Detailed explanation
\(\because \mathrm{P}\left(\mathrm{A} \cap \mathrm{B}^{\mathrm{c}} \cap \mathrm{C}^{\mathrm{c}}\right)=\frac{1}{4}\) Let \(P(A)=x, P(B)=y, P(C)=z\) Then \(x \times(1-y) \times(1-z)=\frac{1}{4}\) ...(i) Similarly, \(P\left(A^c \cap B \cap C^c\right)=\frac{1}{8}\)…
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