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If E1 and E2 are independent events such that P(E1) = 0.3 and P(E2) = 0.4, find

(i) P(E1∩ E2)

(ii) P(E1∩ E2)

(iii) P(\(\overline E_1\cap \overline E_2\) )

(iv) P(\(\overline E_1\cap E_2\))

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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 \(\cup\) E2) when E1 and E2 are independent.

We know that,

Hence, P(E1 \(\cup\) E2) = P(E1) + P(E2) - P(E1 \(\cup\) E2)

= 0.3 + 0.4 – (0.3 x 0.4)

= 0.58

Therefore , P(E1 \(\cup\) E2) = 0.58 when E1 and E2 are Independent.

(iii) P (\(\overline E_1\cap \overline E_2\)) = P(\(\overline E_1\)) x P(\(\overline E_2\))

since , P(E1) = 0.3 and P(E2) = 0.4

\(\Rightarrow\) P(\(\overline E_1\)) = 1 - - P(E1) = 0.7 and P(\(\overline E_2\)) = 1 - P(E2) = 0.6

Since, E1 and E2 are two independent events

\(\Rightarrow\) \(\overline E_1\) and \(\overline E_2\) are also independent events.

Therefore, P(\(\overline E_1\cap \overline E_2\)) = 0.7 x 0.6 = 0.42

(iv) P(\(\overline E_1\) \(\cap\) E2) = P(\(\overline E_1\)) x P(E2)

= 0.7 x 0.4

= 0.28

Therefore , P(\(\overline E_1\) \(\cap\) E2) = 0.28

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