Two Cards Are Drawn One By One From A Pack Of Cards The Probability Of Getting First Card An Ace And Second A Colored One Is 1) 1/26 2) 5/52 3) 5/221 4) 4/13 Solution: Option (3) 5/221 Let E1 be the event of getting an ace. P(E1) = 4/52 = 1/13 P(E2/E1) =... View Article
For Two Events A And B If Pa Pabyb 14 And Pbbya 12 Then 1) A and B are independent 2) P(A’/B) = 3/4 3) P(B’/A’) = 1/2 4) All of the above Solution: Option (4) All of the above (1) P(B/A)... View Article
A And B Are Two Events Such That Pa 08 Pb 06 And Pa Intersection B 05 Then The Value Of Pabyb Is 1) 5/6 2) 5/8 3) 9/10 4) None of these Solution: Option (A) 5/6 Given that, P(A) = 0.8, P(B)= 0.6, and P(A ∩B) = 0.5 P(A/B)... View Article
A Coin Tossed Three Times In Succession If E Is The Event That There Are At Least Two Heads F Is The Event In Which First Throw Is A Head Then Pef 1) 3/4 2) 3/8 3) 1/2 4) 1/8 Solution: Option (1) 3/4 If a coin is tossed, the sample space = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT... View Article
In An Entrance Test There Are Multiple Choice Questions There Are Four Possible Answers To Each Question Of Which One Is Correct The Probability 1) 37/40 2) 1/37 3) 36/37 4) 1/9 Solution: Option (B) 1/37 Let X be the event that the student knows the answer and Y be the event... View Article
In A Certain Town 40 Of The People Have Brown Hair 25 Have Brown Eyes And 15 Have Both Brown Hair And Brown Eyes If A Person Selected 1) 1/5 2) 3/8 3) 1/3 4) 2/3 Solution: Option (3) ⅜ Let A be the event that the people have brown hair and B be the event that the... View Article
The Die Is Tossed And You Are Told That Either Face 1 Or 2 Has Turned Up Then The Probability That Is Face 1 Is Face 1 2 3 4 5 6 Probability 0.1 0.32 0.21 0.15 0.05 0.17 1)... View Article
If C And D Are Two Events Such That C D And Pd Notequals 0 Then The Correct Statement Among The Following Is 1) P(C|D)= P(C) 2) P(C|D)≥ P(C) 3) P(C|D)< P(C) 4) P(C|D)= P(D)/P(C) Solution: Option (2) P(C|D)≥ P(C) We know that P(C|D) =... View Article
A Letter Is Known To Have Come Either From London Or Clifton On The Postmark Only The Two Consecutive Letters On Are Eligible The Probability 1) 5/17 2) 12/17 3) 17/30 4) 3/5 Solution: Option (2) 12/17 Let X be the event that selects a pair of letters from LONDON, and Y be... View Article
If Two Events A And B Are Such That Pabar 03 Pb 04 Pabbar 05 Then Pb By Aubbar Is Equal To 1) 1/2 2) 1/3 3) 1/4 4) None of these Solution: Option (3) ¼ Given that, P(A’) = 0.3, P(B) = 0.4, P(AB’) = 0.5 P(B/(AUB’)) =... View Article
A Biased Die Is Tossed If An Even Face Has Turned Up Then The Probability That It Is Face 2 Or Face 4 Is Face 1 2 3 4 5 6 Probability 0.1 0.24 0.19 0.18 0.15 0.14 1)... View Article
If A And B Are Two Independent Events Such That Pa 12 Pb 15 Then 1) P(A/B) = 1/2 2) P(A/AUB) = 5/6 3) P(A∩B/A’UB’)=0 4) All of these Solution: Option (4) All of these Given that, P(A) = ½, P(B) =... View Article
The Greatest Value Of N For Which 1 Plus 1 By 2 Plus 1 By 2 2plus 1 By 2 N Less Than 2 Is The greatest value of n for which 1 + 1/2 + 1/22 +...+1/2n < 2 is (n ∈N) (1) 100 (2) 10 (3) 1000 (4) none of these Solution: 1... View Article
If Sec X Y Sec X And Sec Xplusy Are In Ap Then Cos X Sec Y By 2 Equal (1) ±√2 (2) ±1/√2 (3) ±2 (4) ±1/2 Solution: Given sec (x-y), sec x and sec (x+y) are in AP. So 2 sec x = sec (x-y) + sec (x+y)... View Article
1 By 2 Plus 3 By 4 Plus 7 By 8 Plus 15 Y 16 Plus N Terms Equal 1/2 + 3/4 + 7/8 + 15/16 +...n terms = (1) n+2-n-1 (2) 2-n-n +1 (3) (2-n-n +1)/4 (4) 2-n +n2 -1 Solution: Let S = 1/2 + 3/4 + 7/8 + 15/16... View Article
If 1 By A 1 By B 1 By C Are In Ap Then 1 By A Plus 1 By B 1 By C 1 By B Plus 1 By C 1 By A Equal (1) (4b2-3ac)/abc (2) 4/ac - 3/b2 (3) 4/ac - 5/b2 (4) (4b2+ 3ac)/ab2c Solution: Given 1/a, 1/b, 1/c are in AP. 1/a - 1/b = 1/b -... View Article
The Sum Of The Series 3 By 4 Plus 5 By 36 Plus 7 By 144 Upto 11 Terms Is (1) 120/121 (2) 143/144 (3) 1 (4) 144/143 Solution: Let S = 3/4 + 5/36 + 7/144 +...upto 11 terms 3/4 = (22-12)/22.12 5/36 =... View Article
6th Term Of The Sequence 7 By 3 35 By 6 121 By 12 335 By 24 Is (1) 2113/96 (2) 2112/96 (3) 2111/96 (4) 865/48 Solution: Given sequence 7/3, 35/6, 121/12, 335/24,... = 2+1/3 , 6-1/6, 10+1/12, …... View Article
If The Sides Of A Triangle Abc Are In Ap And The Greatest Angle Is Double The Smallest The Ratio Of The Sides Of Abc Is (1) 3 : 4 : 5 (2) 5 : 12 : 13 (3) 4 : 5 : 6 (4) 5 : 6 : 7 Solution: Let a,b,c are the sides of the triangle in AP. Let C is the... View Article
1 Plus 4 By 5 Plus 7 By 52 Plus 10 By 53 Upto Infinity Equal (1) 16/35 (2) 11/8 (3) 35/16 (4) 7/16 Solution: Let S = 1+ 4/5 + 7/52 + 10/53 +...upto ∞ …(i) S/5 = 1/5 + 4/52 + 7/53 + 10/53... View Article