1
$\begingroup$

Consider I have the following probabilities:

$$P(A|B) = 0.86 $$

$$ P(A|B^C) = 0.35 $$

$$ P(B) = 0.80 $$

$$ P(A) = 0.758$$

Is there necessary information given to calculate $P(B^C|A^C)$? If so please guide me how. Thanks in advance.

$\endgroup$
  • 1
    $\begingroup$ Is this homework? $\endgroup$ – Sven Hohenstein Apr 18 '13 at 8:25
  • $\begingroup$ Nope, am revising for a test. $\endgroup$ – kype Apr 18 '13 at 9:47
5
$\begingroup$

The formula for the conditional probability is simply: $$ P(B^c\mid A^c)=\frac{P(B^c\cap A^c)}{P(A^c)}. $$ You can calculate the denominator based on your information, so we only need to treat the numerator. Using the formula above with $B^c$ and $A^c$ interchanged, you obtain $$ P(B^c\cap A^c)=P(A^c\mid B^c)P(B^c). $$ Now, try to see if you can find an expression of $P(A^c\mid B^c)$ in terms of $P(A\mid B^c)$.

$\endgroup$
  • $\begingroup$ Thanks for the direction. However way i look at it, even applying bayes theorm I still need this P(B^c|A^c) to calculate this P(A^c∣B^c), which led me to the conclusion that something is missing in the q... $\endgroup$ – kype Apr 18 '13 at 14:15
  • $\begingroup$ You don't need to use Bayes' theorem to find $P(A^c\mid B^c)$. Simply write out the expression and use the law of total probability. $\endgroup$ – Stefan Hansen Apr 18 '13 at 14:36
  • $\begingroup$ Ok what ive got... Complement of a conditional 'P(A^c|B) = 1 - P(A|B) = 0.14' Law of total probability $\endgroup$ – kype Apr 19 '13 at 11:08
  • $\begingroup$ Ok what ive got... Complement of a conditional: P(A^c|B) = 1 - P(A|B) = 0.14 Law of total probability: P(A^c) = P(B) * P(A^c|B) + P(B^c) * P(A^c|B^c) 0.242 = 0.80 * 0.14 + 0.2 * P(A^c|B^c) P(A^c|B^c) = 0.65 Hence P(B^c∩A^c) = 0.65 * 0.2 = 0.13 Which is a wrong answer? So where did i go wrong? Sorry for trouble, i cant edit the comments to look nice $\endgroup$ – kype Apr 19 '13 at 11:17
  • $\begingroup$ @kype: Why not use your first formula for $P(A^c\mid B^c)$? Then $P(A^c\mid B^c)=1-P(A\mid B^c)$ which you can compute. $\endgroup$ – Stefan Hansen Apr 19 '13 at 11:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.