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Addition Theorems on Probability




Addition Theorems on Probability

  1. Let E and F are two events  of an random experiments
    Then
    $P(E \cup F) = P(E) + P(F) - P(E \cap F) $
    I event E and F are mutually exclusive
    $n( E \cap F) =0$
    Then
    $P(E \cup F) = P(E) + P(F)$
  2. Addition Theorem of three events
    Let E ,F and G are three events  of an random experiments
    $P(E \cup F \cup G ) = P(E) + P(F) + P(G) - P(E \cap F) -  P(F \cap G) - P(E \cap G )  + P(E \cap F \cap G)$
    If event E , F and G are mutually exclusive
    $n( E \cap F) = 0$
    $n(E \cap G) = 0$
    $n(F \cap G) = 0$
    $n(E \cap F \cap G)= 0$
    Then
    $P(E \cup F\cup G ) = P(E) + P(F) + P(G)$
  3. Let E and F are two events  of an random experiments
    Then
    Probability of the occurrence of the Event E only $( E \cap F^c)$
    $ P(E \cap F^c) = P(E) - P(E \cap F )$
    Probability of the occurrence of the Event F only $( E^c \cap F)$
    $P(E^c \cap F) = P(F) - P(E \cap F )$
    Probability of occurrence of exactly one of the two event E and F
    $P[(E^c \cap F) \cup (E \cap F^c) ] = P(E) + P(F) - 2 P(E \cap F)$



Solved examples

Example 1
E, F and G are events associated with a random experiment such that
$P(E) = 0.3, P(F) = 0.4, P(G) = 0.8$ $P(E \cap F) = 0.08 ,P( E \cap G) = 0.28 \; and \; P(E \cap F \cap G) = 0.09$. If $P(E \cup F \cup G) \geq 0.75$ then prove that $P(F \cap G)$ lies in the interval [0.23, 0.48]
Answer
We know that
$P(E \cup F \cup G ) = P(E) + P(F) + P(G) - P(E \cap F) $
$- P(F \cap G) - P(E \cap G ) + P(E \cap F \cap G)$
$P(E \cup F \cup G ) = .3 +.4+.8 -.08-.28 +.09 - P(F \cap G)$
$P(E \cup F \cup G ) =1.23 - P(F \cap G)$ -(1) Or
$1.23 - P(F \cap G) \geq 0.75$
$1.23 -.75 \geq P(F \cap G)$
$.48 \geq P(F \cap G) $
Now we know any Probability is less than 1
So
$P(E \cup F \cup G ) \leq 1$
So
$1.23 - P(F \cap G) \leq 1$
Then
$P(F \cap G) \geq .23$
So
$.23 \leq P(F \cap G) \leq .48$

Example 2
From a pack of 52 cards, two cards are drawn together, what is the probability that both the cards are kings
Answer
Total outcome = 52C2  = 1326
Total King cases =4C2 =6
Probability ==6/1326=1/221

Example 3
Three unbiased coins are tossed. What is the probability of getting at most two heads?
Answer
S={TTT,TTH,THT,HTT,THH,HTH,HHT,HHH}
Let E be the event of getting at most two heads
E={TTT,TTH,THT,HTT,THH,HTH,HHT}
P(E) =7/8

Example 4
Find the probability of selecting a red card or a 7 from a deck of 52 cards?
Answer
We need to find out P(R or 7)
Probability of selecting a RED card=P(R) = $\frac {26}{52}$
Probability of selecting a 7 =P(7) = $\frac {4}{52}$
Probability of selecting both a red card and a 7
$P( R \cap 7) = \frac {2}{52}$
Now $P(R \cup 7) = P(R) + P(7) - P(R \cap 7)$
$= \frac {26}{52} + \frac {4}{52} - \frac {2}{52}$
$= \frac {28}{52}$

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