Answer :

Given that P(E1) =P1 and P(E2) =P2


(i)P1P2


=P(E1). P(E2) =P (E1 Ո E2)


So, E1 and E2 occur simultaneously.


(ii)(1-P1) P2


As we know P(A)+P(A) = 1


=P(E1)’. P(E2)


= P (E1’ Ո E2)


So, E1 does not occur but E2 occur.


(iii)1-(1-P1) (1-P2)


As we know P(A)+P(A) = 1


=1-P(E1)’P(E2)’


=1-P (E1’ Ո E2’)


By De Morgan’s laws:



=1-P (E1 Ս E2)


As we know P(A)+P(A) = 1


= 1- [1-P (E1 Ս E2)]


= P (E1 Ս E2)


So, either E1 or E2 or both E1 and E2 occurs.


(iv)P1+P2-2P1P2


= P(E1) + P(E2) – 2P(E1) P(E2)


= P(E1) + P(E2) – 2P (E1 Ո E2)


= P (E1 Ս E2)- 2P (E1 Ո E2)


So, either E1 or E2 occurs but not both.


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