- 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

$P(E \cup F) = P(E) + P(F)$ - 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

$P(E \cup F\cup G ) = P(E) + P(F) + P(G)$ - 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

$P[(E^c \cap F) \cup (E \cap F^c) ] = P(E) + P(F) - 2 P(E \cap F)$

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]

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

$.23 \leq P(F \cap G) \leq .48$

From a pack of 52 cards, two cards are drawn together, what is the probability that both the cards are kings

Total outcome =

Total King cases =

Probability ==6/1326=1/221

Three unbiased coins are tossed. What is the probability of getting at most two heads?

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

Find the probability of selecting a red card or a 7 from a deck of 52 cards?

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 $P(R \cup 7) = P(R) + P(7) - P(R \cap 7)$

$= \frac {26}{52} + \frac {4}{52} - \frac {2}{52}$

$= \frac {28}{52}$

Class 12 Maths Class 12 Physics

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