Question

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(\(\overline E_1\cap E_2\))

Answer 1

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