GO Classes 2024 | Weekly Quiz 8 | Conditional Probability | Question: 3

Answer 1

We have sample space $S$.

$k$ events $B_{1},B_{2},…,B_{k}$ form a partition of sample space $S$. There are another event $A$ have $Pr(A)>0$.

Now, according to $axioms$ $of$ $probability$ we can say that, if all of these events are $disjoint$ then we can write like this:

$P(B_{1} \cup B_{2}, … \cup B_{k}) = P(B_{1}) + P(B_{2}) + … + P(B_{k})$

So, according to our $2^{nd}$ $axiom$: $P(B_{1} \cup B_{2}, … \cup B_{k}) = 1$

Then also according to our $3^{rd}$ $axiom$ we can say that, 

$P(B_{1}) + P(B_{2}) + … + P(B_{k}) = 1$

$\therefore$ $\sum_{i=1}^{k} P(B_{i}) = 1$

Now, Conditional probability also satisfies the $axioms$ like probability.

So, we can say that if there $A$ event also given to us then also we can get total probability summation as $1$ for the given sample space. $(Pr(A)>0)$ $[given]$

$\therefore$ $\sum_{i=1}^{k} P(B_{i} | A) = 1$  this also $valid$.

$Partition$ $Theorem$ $/$ $Total$ $Probability$ 

Now by these we know that if $B_{1},B_{2},…,B_{k}$ making partition in sample space for any event $A$ then,

$P(A) = \sum_{i=1}^{k} P(B_{i} \wedge A)$ $= P(A \wedge B_{1}) + P(A \wedge B_{2}) + … P(A \wedge B_{k})$

$Ans: A;B;D$

Some Resources: 

$University$ $of$ $Auckland$ handbook: https://www.stat.auckland.ac.nz/~fewster/325/notes/ch2annotated.pdf [Pg. No. 23 – 26]

Comments

hey thanks akash for the handbook, i was wondering how to solve this question

Answer 2

Suppose that k events B1,B2,B3,…,Bk form a partition of the sample space S .(B1 U B2 U B3 U...U Bk)=S    =>   P(B1 U B2 U B3 U...U Bk)=1   =>

 P(B1)+P(B2)+P(B3)+...+P(Bk)=1  =>  Σ i=1 to i=k P(Bi)=1 

now P(A)=P(B1 ∩ A)+P(B2∩A)+...+P(Bk∩A)=Σ i=1 to i=k P(Bi∩A)

               =P(B1|A)*P(A)+P(B2|A)*P(A)+...+P(Bk|A)*P(A)= P(A) *Σ i=1 to i=k P(Bi|A)

So Σ i=1 to i=k P(Bi|A)=1 

So A,B,D are the correct options .