Given: E1 and E2 are two independent events such that P(E1) = 0.3 and P(E2) = 0.4
To Find:
(i) P(E1 ∩ E2)
We know that,
when E1 and E2 are independent ,
P(E1 ∩ E2) = P(E1) x P(E2)
= 0.3 x 0.4
= 0.12
Therefore, P(E1 ∩ E2) = 0.12 when E1 and E2 are independent.
(ii) P(E1 ∪
E2) when E1 and E2 are independent.
We know that,
Hence, P(E1 ∪
E2) = P(E1) + P(E2) - P(E1 ∪
E2)
= 0.3 + 0.4 – (0.3 x 0.4)
= 0.58
Therefore , P(E1 ∪
E2) = 0.58 when E1 and E2 are Independent.
(iii) P (
¯
E
1∩
¯
E
2
) = P(
¯
E
1
) x P(
¯
E
2
)
since , P(E1) = 0.3 and P(E2) = 0.4
⇒
P(
¯
E
1
) = 1 - - P(E1) = 0.7 and P(
¯
E
2
) = 1 - P(E2) = 0.6
Since, E1 and E2 are two independent events
⇒
¯
E
1
and
¯
E
2
are also independent events.
Therefore, P(
¯
E
1∩
¯
E
2
) = 0.7 x 0.6 = 0.42
(iv) P(
¯
E
1
∩
E2) = P(
¯
E
1
) x P(E2)
= 0.7 x 0.4
= 0.28
Therefore , P(
¯
E
1
∩
E2) = 0.28
