Events `Aa n dC` are independent. If the probabilities relating `A ,B ,a n dC` are `P(A)=1//5, P(B)=1//6; P(AnnC)=1//20 ; P(BuuC)=3//8.` Then events `

Monjur Alahi

Events `Aa n dC` are independent. If the probabilities relating `A ,B ,a n dC` are `P(A)=1//5, P(B)=1//6; P(AnnC)=1//20 ; P(BuuC)=3//8.` Then events `Ba n dC` are independent events `Ba n dC` are mutually exclusive events `Ba n dC` are neither independent nor mutually exclusive events `Ba n dC` are equiprobable
A. events B and C are independent
B. events B and C are mutually exclusive
C. events B and C are neiter independent nor mutually exclusive
D. events B and C are equiprobabile


0 like 0 dislike
1 views
Share with your friends

1 Answers

Salim Ahmed Kawsar

Call

Correct Answer - A
`P(AnnC)=P(A)P(C)`
`or1/20=1/5P(C)`
`orP(C)=1/4`
Now, `P(BuuC)=1/6+1/4-P(BnnC)`
`P(BnnC)=5/12-3/8=1/24P(BnnC)`
Therefore, B and C aer independent.

0 like 0 dislike
1 views
Talk Doctor Online in Bissoy App
If two events are independent, then

If two events A and B are independent, then we knpw that `P(AcapB)=P(A)cdotP(B),P(A)ne0,P(B)ne0` `Since,m a nd B have a common outcome Further, mutually exlusive events never have a common outcome....

2 Answers 1 views
If the events A and B are independent, then `P(AcapB)` is equal to

If A and B are independent, then `P(AcapB)=P(A)cdotP(B)`

2 Answers 1 views
Two events E and F are independent. If P(E )=0.3 and P(`EcupF`)=0.5 then `P(E//F)-P(F//E)` equals to

Here, P(E )=0.3 and `P(EcupF)=0.5` Let P(F)=x ltbr gt`becauseP(EcupF)=P(E )+P(F)-P(EcapF)` `=P(E )+P(F)-P(E )cdotP(F)` `rArr0.5=0.3+x-0.3x` `rArrx=(0.5-0.3)/0.7=2/7=P(F)` `thereforeP(E//F)-P(F//E)=(P(EcapF))/(P(F))-(P(FcapE))/(P(E ))` `=(P(EcapF)cdotP(E)-P(FcapE)cdotP(F))/(P(E)cdotP(F))` `=(P(EcapF)[P(E)-P(F)])/(P(EcapF))=P(E)-P(F)` `=3/10-2/7=(21-20)/70=1/70`

2 Answers 1 views
The probabilities of three events `A ,B ,a n dC` are `P(A)=0. 6 ,P(B)=0. 4 ,a n dP(C)=0. 5.` If `P(AuuB)=0. 8 ,P(AnnC)=0. 3 ,P(AnnBnnC)=0. 2 ,a n dP(A

We have, `P(A nn B) = P(A) + P(B) - P(A uu B) = 0.6 + 0.4 - 0.8 = 0.2` Also, `P(A uu B uu C) = P(A) +...

2 Answers 1 views
Let `A ,B ,C` be three events such that `P(A)=0. 3 ,P(B)=0. 4 ,P(C)=0. 8 ,P(AnnB)=0. 88 ,P(AnnC)=0. 28 ,P(AnnBnnC)=0. 09.` If `P(AuuBuuC)geq0. 75 ,` t

Since `P(A uu B uu C) ge 0.75`, we have `0.75 le P(A) + P(B) + P(C ) - P(A nn B) - P(B nn C) - P(A nn C)...

2 Answers 1 views
Let A,B, C be three mutually independent events. Consider the two statements `S_(1)and S_(2).` `{:(S_(1):A and B nnC "are independent.",),(S_(2):A and

Correct Answer - A We are given that `P(AnnB)=P(A)P(B)` `P(BnnC)=P(B)P(C)` `P(CnnA)=P(C)P(A)` `P(AnnBnnC)=P(A)(B)P(C)` We have, `P(Avv(BnnC)=P(A)(B)P(C)=P(A)P(B)P(C)=P(A)P(BnnC)` `impliesA and BnnC` are independent Therefore, `S_(2)` is true. Also, `P[(Ann(BuuC)]=P[(AnnB)uu(AnnB)nn(AnnC)]` `=P(AnnB)+P(AnnC)-P(AnnBnnC)` `=P(A)P(B)+P(A)P(C)-P(A)P(B)P(C)` `P(A)[P(B)+P(C)-P(B)P(C)]` `P(A)[P(B)+P(C)-P(BnnC)]` `=P(A)P(BnnC)`

2 Answers 1 views
Let S and T are two events difined on a sample space with probabilities `P(S)=0.5,P(T)=069,P(S//T)=0.5` Events S and T are

Correct Answer - B `P(S//T)=(P(SnnT))/(P(T))` `or 0.5=(P(SnnT))/(0.69)` ` or 0.5=(P(SnnT))/(0.69)` `or P(SnnT)=0.5xx0.69=P(S)P(T)` Therefore, S and T are independent. `thereforeP(Aand T)=P(S)P(T)` ` =0.69xx0.5=0.345` `P(SorT)=P(S)+P(T) -P(SnnT)` `=0.5+0.69-0.345 =08450`

2 Answers 1 views
Let S and T are two events difined on a sample space with probabilities `P(S)=0.5,P(T)=069,P(S//T)=0.5` The value of P(S and T) is

Correct Answer - A `P(S//T)=(P(SnnT))/(P(T))` `or 0.5=(P(SnnT))/(0.69)` ` or 0.5=(P(SnnT))/(0.69)` `or P(SnnT)=0.5xx0.69=P(S)P(T)` Therefore, S and T are independent. `thereforeP(Aand T)=P(S)P(T)` ` =0.69xx0.5=0.345` `P(SorT)=P(S)+P(T) -P(SnnT)` `=0.5+0.69-0.345 =08450`

2 Answers 1 views
Let S and T are two events difined on a sample space with probabilities `P(S)=0.5,P(T)=069,P(S//T)=0.5` The value of P(S or T) is

Correct Answer - C `P(S//T)=(P(SnnT))/(P(T))` `or 0.5=(P(SnnT))/(0.69)` ` or 0.5=(P(SnnT))/(0.69)` `or P(SnnT)=0.5xx0.69=P(S)P(T)` Therefore, S and T are independent. `thereforeP(Aand T)=P(S)P(T)` ` =0.69xx0.5=0.345` `P(SorT)=P(S)+P(T) -P(SnnT)` `=0.5+0.69-0.345 =08450`

2 Answers 1 views
Football teams `T_(1)and T_(2)` have to play two games are independent. The probabilities of `T_(1)` winning, drawing and lossing a game against `T_(2

Correct Answer - B `P(XgtY)=T_(1)T_(1)+DT_(1)+T_(1)D("where"T_(1)" represents wins and D represents draw")` `=((1)/(2)xx(1)/(2))+((1)/(6)xx(1)/(2))+((1)/(2)xx(1)/(6))=5/12`

2 Answers 1 views
X