A box `B_(1)` contains 1 white ball, 3 red balls, and 2 black balls. An- other box `B_(2)` contains 2 white balls, 3 red balls. A third box `B_(3)` co

Here, `W_(1)` = {4 white balls} and `B_(1)`={5 black balls} and `W_(2)` = {9white balls} and `B_(2)` = {7 black balls} Let E, is the event that ball transferred fram...

Bag I ={3B,2W],Bag II={2B,4W} Let `E_(1)`=Event that bag I is selected `E_(2)`=Event that bag II is selected and E=Event that a black ball is selected `rArrP(E_(1))=1//2,P(E_(2))=1/2,P(E//E_(1))=3/5,P(E//E_(2))=2/6=1/3` `thereforeP(E)=P(E_(1))cdotP(E//E_(1))+P(E_(2))cdotP(E//E_(2))` `1/2cdot3/5+1/2cdot2/6=3/10+2/12` `=(18+10)/60=28/60=7/15`

Bag I : 3 red balls and 0 white ball Bag II : 2 red balls and 1 white ball. Bag III : 0 red ball andf 3 white balls....

Since, Bag I= {3 black, 4 white balls}, Bag II = {4 black, 3 white balls} Let `E_(1)` be the event that bag is selected and `E_(2)` be the event...

Let `U_(1)`={2 white, 3 black balls} `U_(2)`={3 white , 2 black balls} and `U_(3)`={4 white ,1 black balls} `thereforeP(U_(1))=P(U_(2))=P(U_(3))=1/3` Let `E_(1)` be the event that a ball is chosen from...

Procedure of drawing th balls has to end at the rth draw. So, exactly two black balls are drawn in first (r-1) draws and third black ball is drawn in...

Correct Answer - `1//4` If third ball is red, then in first two draws, there will be either no red ball or one red ball. Let events `R_(0)=` in first two...

Correct Answer - A Consider the following events. A: ball drawn is black `E_(1):` bag I is chosen `E_(2):` bag II is chosen `E_(3):` bag III is chosen Then, `P(E_(1))=(E_(2))=P(E_(3))=1/3` `P(A//E_(1))=3/5,P(A//E_(2))=1/5,P(A//E_(3))=7/10`...

Correct Answer - A::B `Red ton_(1)` ` "Box" IltBlack ton_(2)` `Red to n_(3)` `"Box II"lt"Black" to n_(4)` `P(R)=1/2.(n_(1))/(n_(1)+n_(2))+1/2.(n_(3))/(n_(3)+n_(4))` `P (II//R) =(1/2(n_(3))/(n_(3)+n_(4)))/(1/2(n_(1))/(n_(1)+n_(2))+1/2.(n_(3))/(n_(3)+n_(4)))` `((n_(3))/(n_(3)+n_(4)))/((n_(1))/(n_(1)+n_(2))+(n_(3))/(n_(3)+n_(4)))` by option `n_(1)=3,n_(2)=3,n_(3)=5,n_(4)=15` `P(II//R)=(5/20)/(3/6+5/20)=(1/4)/(1/2+1/4)=1/4xx(4)/(2+1)=1/3` Also, when `n_(1)=3, n_(2),=6,n_(3)=10, n_(4)=50,P(II//R)=1//3`

Correct Answer - B `(b)` Let `A` be the event of drawing a black ball. `A_(1)`, the event of choosing the first box. `A_(2)`, the event of choosing the second box....