The probability of event $A$ occurring is $P(A)=13/25$.The probability of event B acquiring is $P(B) = 9/25$. The conditional probability of $A$ occurring given that $B$ has occurred is $P(A|B) = 5/9$

(a) Determine the following probabilities

i - $P( A$ and $B)$

ii - $P(B|A)$

iii- $P(A$ or $B$ or both)

iv- P(Not A | Not B)

(b) Determine $P( A$ occurs or $B$ does not occur), show your working

I have done the first $3$ parts of the question $18$ (a(i-iii)) but at fourth part of a and part $b$, I wasn't sure how to work it out

The answers are:

ai- $1/5$

aii- $5/13$

aiii- $17/25$

aiv- $1/2$


This is an a level question and any help will be much appreciated.


Hint for a) iv:

$$ P(A^c \mid B^c) = \frac{ P(A^c \cap B^c) }{P(B^c)} = \frac{P( (A \cup B)^c )}{P (B^c)} $$

Hint for b): $$ P(A \cup B^c) = P(A) + P(B^c) - P(A \cap B^c) $$ For that last intersection you can apply the rule that $$ P(A \cap B) = P(A \mid B)P(B) $$ and then look back at a) iv

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  • $\begingroup$ Thanks a lot. I'll give it go $\endgroup$ – Nikhil Patel Jan 22 '17 at 17:15
  • $\begingroup$ I have the answer part a iv , thanks. Still can't get part b as my value for P(A and NotB) seems wrong. I keep getting 8/25. Could I get a bit more help? $\endgroup$ – Nikhil Patel Jan 22 '17 at 19:16
  • $\begingroup$ Never mind got it, thanks for all the help $\endgroup$ – Nikhil Patel Jan 22 '17 at 19:20
  • $\begingroup$ @NikhilPatel glad to help. Remember to mark answers as accepted if you're satisfied. $\endgroup$ – Slug Pue Jan 23 '17 at 8:17

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