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  • If A and B such events that P(A) > 0 and P(B) ≠ 1, then P(A’|B’) equals A. 1 – P(A|B) B. 1 – P (A’|B)

If A and B such events that P(A) > 0 and P(B) ≠ 1, then P(A’|B’) equals
A. 1 – P(A|B)

B. 1 – P (A’|B)

C. \frac{1-\mathrm{P}(\mathrm{A} \cup \mathrm{B})}{\mathrm{P}\left(\mathrm{B}^{\prime}\right)}

D. P(A’) | P(B’)
 

Answers (1)

Given-
\\P(A)>0$ and $P(B) \neq 1$ \\$\mathrm{P}\left(\mathrm{A}^{\prime} \mid \mathrm{B}^{\prime}\right)=\frac{P\left(A^{\prime} \cap \mathrm{B}^{\prime}\right)}{P\left(B^{\prime}\right)}
By de Morgan's Law:
\\P\left(A^{\prime} \cap B^{\prime}\right)=P(A \cup B)^{\prime}$ \\$=\frac{1-P(A \cup B)}{P\left(B^{\prime}\right)}

Option c is correct answer.

Posted by

infoexpert22

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