p(a' ∩ b') formula

MinatoNamikaze

Is there any formula for it?

Reply 1

Guarddyyy

Surely it's just 1 - p(a) - p(b) + p(a and b)?

Reply 2

Notnek

Original post by MinatoNamikaze

Is there any formula for it?

You can make a formula for anything. Draw a Venn diagram and try to come up with one.

Reply 3

MinatoNamikaze

Original post by Notnek

You can make a formula for anything. Draw a Venn diagram and try to come up with one.

Thanks I tried and made two intersecting circles and subtracted bothe the circles from the total and added the intersection as it was overlapped and it worked out just as @guarddyyy said. Thank you for the help :smile:

Reply 4

RDKGames

Study Forum Helper

Original post by MinatoNamikaze

Is there any formula for it?

De Morgan's law is also a thing you might find useful.

P(A' \cap B') = P([A \cup B]') = 1-P(A \cup B) = 1- P(A) - P(B) + P(A \cap B)

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p(a' ∩ b') formula

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