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Independent Events

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Independent Events - Lesson Summary

Events E and F are said to be independent events, if the probability of the occurrence of one event is not affected by the occurrence of the other event.

Two events, E and F, are said to be independent events, if:

• P(E|F) = P(E), where P(F) ≠ 0

• P(F|E) = P(F), where P(E) ≠ 0

By the multiplication rule of probability,

P(E ∩ F) = P(E|F).P(F)

Ex:

In an experiment - draw a card from a deck of 52 playing cards. The card drawn is a heart.

and the card drawn is an ace.

Sol: Each event of the experiment is equally likely.

E: The card drawn is a heart.

F: The card drawn is an ace.

Number of hearts = 13

Number of aces = 4

P(E) = 13/52 = 1/4

P(F) = 4/52 = 1/13

P(E ∩ F) = 1/52 (Since there is only one ace of hearts in a deck of 52 playing cards.)

P(E|F) = P(E∩F)/P(F) = (1/52)/(1/13) = 1/4

P(E|F) = 1/4 = P(E)

P(E) is not affected by the occurrence of event F.

P(F|E) = P(E∩F)/P(E) = (1/52)/(1/4) = 1/13

P(F|E) = 1/13 = P(F)

P(F) is not affected by the occurrence of event E.

Observe that in both the cases, the probability of occurrence of one event is not affected by the occurrence of the other event. Such events are called independent events.

If E and F are independent events, then P(E|F) = P(E).

Events E and F are said to be independent events, if P(E ∩ F) = P(E).P(F).

Two events, E and F, are said to be dependent, if P(E ∩ F) ≠P(E).P(F).

Three events, E, F and G, are said to be mutually independent, if:

• P(E ∩ F) = P(E).P(F)

• P(E ∩ G) = P(E).P(G)

• P(F ∩ G) = P(F).P(G)

• P(E ∩ F ∩ G) = P(E).P(F).P(G)

Note:

Events E, F and G are said to be dependent, if anyone of the conditions is not satisfied.


Properties of Independent Events

1) If E and F are two independent events associated with some experiment, then E and F′ are also independent.

Given: P(E ∩ F) = P(E).P(F)

To verify: P(E ∩ F′) = P(E).P(F′)


Verification:

Form the Venn diagram:

• E ∩ F and E ∩ F′ are mutually exclusive events.

• E = (E ∩ F) ∪ (E ∩ F′) [∵(E ∩ F) ∪ (E ∩ F′) = E ∩ (F ∪ F′)

= E ∩ S = E

P(E) = P(E ∩ F) + P(E ∩ F′)

P(E ∩ F′) = P(E) - P(E ∩ F)

P(E ∩ F′) = P(E) - P(E).P(F) (Since E and F are independent events.)

= P(E)[1 - P(F)]

= P(E).P(F′) [∵ P(F)+ P(F′) = 1 ]

∴ P(E ∩ F′) = P(E).P(F′)

Hence, E and F′ are independent events.

Similarly, E′ and F are independent events, and E′ and F′ are also independent events.


2)If E and F are two independent events associated with some experiment, then the probability of the occurrence of "at least one of E and F" is given by P(E ∪ F) = 1- P(E′) P(F′).

Given: P(E ∩ F) = P(E).P(F)

To verify: P(E ∪ F) = 1- P(E′) P(F′)

Verification:

P(at least one of E and F) = P(E ∪ F)

= P(E) + P(F) - P(E ∩ F)

= P(E) + P(F) - P(E).P(F)

= P(E) + P(F)[1 - P(E)]

= P(E) + P(F).P(E′) [∵ P(E)+ P(E′) = 1 ]

= 1 - P(E′) + P(F).P(E′)

= 1 - P(E′)[1 - P(F)]

= 1 - P(E′).P(F′)

P(E ∪ F) = 1- P(E′) P(F′)

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