# 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)

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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