`{:(,"ColumnI",,"ColumnII"),((A),(K.E.)/(P.E.),(P),2),((B),P.E+2K.E.,(Q),-(1)/(2)),((C),(P.E.)/(T.E.),(R),1),((D),(K.E.)/(T.E.),(S),0):}`

If E and F are the complementary events of events E and F, respectively,and if 01, then P((E)/(F))+P((E)/(F))=(1)/(2)bP((E)/(F))+P((E)/(F))=1cP((E)/(F))+P((E)/(F))=(1)/(2)dP((E)/(F))+P((E)/(F))=1

Relation between kinetic energy and momentum Let us consider a body of mass 'm' having a velocity 'v', then momentum of the body P = mass xx velocity P = m xx v rArr v = (P)/(m) " "...(1) From definition, kinetic energy (K.E) of the body K.E = (1)/(2) mv^(2)" "...(2) Now putting the value of (1) in (2) we have K.E = (1)/(2)m ((P)/(m))^(2) K.E. = (1)/(2)m (P^(2))/(m^(2)) = (1)/(2) (P^(2))/(m) = (P^(2))/(2m)" "...(3) Thus we can write P^(2) = 2m xx K.E rArr P = sqrt(2m xx K.E) Thus momentum = sqrt(2 xx "mass" xx "kinetic energy") The kinetic energy of a given body is doubled. Its momentum will

If E_(1) and E_(2) are events of a sample space such that P(E_(1))=(1)/(4),P((E_(2))/(E_(1)))=(1)/(2),P((E_(1))/(E_(2)))=(1)/(4) Then P((E_(1))/(E_(2)))=

if E_(1) and E_(2) are two events such that P(E_(1))=(1)/(4),P((E_(2))/(E_(1)))=(1)/(2) and P=((E_(1))/(E_(2)))=(1)/(4)

Suppose E_(1) and E_(2) are two events of a random experiment such that P(E_(1))=(1)/(4), P((E_(2))/(E_(1)))=(1)/(2) and P((E_(1))/(E_(2)))=(1)/(4) then P(E_(2))=

Let E^(c) denote the complement of an event E . Let E,F,G be pairwise independent events with P(G)>0 and P(E nn F nn G)=0 Then P(E^(c)nn F^(c)nn G) equals (A)P(E^(c))+P(F^(c)) (B) P(E^(c))-P(F^(c))(C)P(E^(c))-P(F)(D)P(E)-P(F^(c))

A sample space consists of 9 elementary outcomes E_(1),E_(2),…..,E_(9) whose probabilities are P(E_(1))=P(E_(2))=0.08,P(E_(3))=P(E_(4))=P(E_(5))=0.1 P(E_(6))=P(E_(1))=0.2,P(E_(8))=P(E_(9))=0.07 "Suppose"" "A={E_(1),E_(5),E_(8)},B={E_(2),E_(5),E_(8),E_(9)} (i) Calculate P(A), P(B) and P(AcapB) . (ii) Using the addition law of probability, calculate P (AcupB). (iii) List the composition of the event AcupB and calculate P(AcupB) by adding the probabilities of the elementary outcomes. Calculate P(overset(-)B) from P(B), also calculate P(overset(-)B) directly from the elementary outcomes of overset(-)B ,

If E_(1) nad E_(2) are two independent events such that P(E_(1))=0.3 and P(E_(2))=0.6 then find the value of (i)P(E_(1)nn E_(2)) (ii) P(E_(1)uu E_(2)) (iii) P((E_(1))/(E_(2)))( iv )P((E_(2))/(E_(1)))