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Probability Some Rules Of Probability L-3
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Probability
lecture-3
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Probability Some Rules Of Probability L-3
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Axioms of probability
- The following three axioms - 1. P(E) $\geq$ 0 for all E $\in$ C - 2. P(S) = 1 - 3. let E_1, E_2,.... be pairwise disjoint events then - $P(\bigcup_{i=1}^{\infty} E_i)=\sum_{i=1}^{\infty} P\(E_i)$ - a,b,c are disjoint events - $ P(A \cup B)=P(A)+P(B) $ - $ P(A \cup B \cup C)=P(A)+P(B)+P(C)$
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Probability Some Rules Of Probability L-3
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Addition rule of probability
- Let $A$ and $B$ be any two events. Then - $P(A \cup B)=P(A)+P(B)-P(A \cap B)$ - Proof: $ ~ $ We can write - $A \cup B=A \cup(B-(A \cap B))$ - So $P(A \cup B)= P(A) + P(B-(A \cap B)) $ - = $ P(A)+P(B)-P(A \cap B)$
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Probability Some Rules Of Probability L-3
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Probability a monotonic function
- $E= F \cup(E-F) $ - $P(E)= P(F)+P(E-F) $ - So $ P(E-F)=P(E)-P(F) \geq 0 $ - $\Rightarrow P(E) \geq P(F) $ - i.e probability is a monotone function.
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Probability Some Rules Of Probability L-3
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$P(E^c)$
- 3) $E \cup E^c=S $ - $P(E)+P\left(E^c\right)=P(S)=1 $ - $\Rightarrow P\left(E^c\right)=1-P(E) $
![alt text](https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/9a8b9a22-44a1-4241-1df8-67a0d06d0000/public)
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Probability Some Rules Of Probability L-3
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Addition rule of probability extension to 3 events
- Extension to 3 events: $A, B, C$ - $ P(A \cup B \cup C)=P(A \cup B)+P(C)-P((A \cup B) \cap C) $ - $ =P(A)+P(B)-P(A \cap B)+P(C) $ - $ \quad-P((\underbrace{A \cap C)} \cup(B \cap C)) $ - $ =P(A)+P(B)+P(C)-P(A \cap B) $ - $ \quad-\quad(P(A \cap C)+P(B \cap C)-P((A \cap C) \cap(B \cap C))) $ - $ =P(A)+P(B)+P(C)-P(A \cap B)-P(A \cap C) $ - $ \quad-P(B \cap C)+P(A \cap B \cap C)$
![alt text](https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/09175988-e506-41fa-4485-8f805a9f2d00/public)
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Probability Some Rules Of Probability L-3
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General addition rule
- Let $A_1, A_2, \ldots, A_n$ be any events. Then - $P(\bigcup_{i=1}^n A_i)=\sum_{i=1}^n P(A_i)-\underset{i
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Probability Some Rules Of Probability L-3
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General addition rule proof
- Proof: We will prove the relation (1) using the principle of mathematical induction. - P(1) - $P(K) \rightarrow P(K+1)$ - For $n=1$, the statement (1) becomes - $P(A_1)=P(A_1)$ which always holds true. - Next we assume the statement (1) to be true for $n=k$. Then we prove it for $n=k+1$
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Probability Some Rules Of Probability L-3
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General addition rule proof
- $ P(\bigcup_{i=1}^{k+1} A_i)=P((\underbrace {\bigcup_{i=1}^k A_i}) \cup A_{k+1}) $ - $ = P(\bigcup_{i=1}^k A_i)+P(A_{k+1})-P((\bigcup_{i=1}^k A_i) \cap A_{k+1})$ - (by addition rule for two events) - $ =\sum_{i=1}^k P(A_i)-\underset{i
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Probability Some Rules Of Probability L-3
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General addition rule proof
- $ =\sum_{i=1}^{k+1} P(A_i) - \underset{i
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Probability Some Rules Of Probability L-3
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General addition rule proof
- $= \sum_{i=1}^{k+1} P(A_i)- \underset{i
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Probability Some Rules Of Probability L-3
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Problem
- Suppose six cards are drawn one by one with replacement from a well shuffled pack if 52 cards. We want bo find out the probability that in this set of six cards each of the four suits (heart, spades club, diamond) appear.
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Probability Some Rules Of Probability L-3
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Solution
- (Using the general addition ). - Let A denote the event that the set of six cards contains at least one card of each suit . - Then $A^c \rightarrow$ at least one suit does not appear in the set $\eta$ six cards. - Let, - $B_1 \rightarrow$ hearts do not appear - $B_2 \rightarrow$ spades do not appear - $B_3 \rightarrow$ clubs do not appear - $B_4 \rightarrow$ diamonds do not appear - Then $ A^c=\bigcup_{i=1}^4 B_i$
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Probability Some Rules Of Probability L-3
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Solution
- By general addition rule - $P(A^c) = P(\bigcup_{i=1}^4 B_i) $ - $= \sum_{i=1}^4 P(B_i) - \underset{i
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Probability Some Rules Of Probability L-3
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Solution
- $ P(B_i \cap B_j \cap B_k) = (\frac{1}{4})^6, \quad (1 \leq i
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Probability Some Rules Of Probability L-3
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