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If E and F are events such that P(E) =1/4 , P(F) =1/2 and P(E and F) =1/8, find:(i) P(E or F), (ii) P(not E and not F).
Concept: Probability - Probability of 'Not', 'And' and 'Or' Events
Is there an error in this question or solution?
Text Solution
Answer : N/a
Solution : Here `P(E)=1/4, P(F)=1/2` and `P(EnnF)=1/8` <br> (i) `P(E "or"F)=P(EuuF)` <br> `=P(E)+P(F)-P(EnnF)` <br> `=1/4+1/2-1/8` <br> `=(2+4-1)/8=5/8` <br> (iii) `P` (E -not and F-not) `=P(E'nF')=P(EuuF)'` <br> `=1-P(EuuF)` <br> `=1-5/8=3/8`
Text Solution
Solution : We have given, P(E)= `1/4` , P(F)= `1/2` and P(E and F) = `1/8` <br> (E or F) = P(E) + P(F) - P(E and F) <br> = `1/4` - `1/2` - `1/8` = `5/8`