A Bag X Contains 2 White And 3 Black Balls And Another Bag Y Contains 4 White And 2 Black Balls The Probability For The Ball Chosen Be White Is 1) 2/15 2) 7/15 3) 8/15 4) 14/15 Solution: Option (3) 8/15 Let E1 be the event that the ball is drawn from bag “X” and E2 be the... View Article
A Bag A Contains 2 White And 3 Red Balls And Bag B Contains 4 White And 5 Red Balls One Ball Is Drawn Is Found To Be Red From Bag B Was 1) 5/14 2) 5/16 3) 5/18 4) 25/52 Solution: Option (4) 25/52 Let E1 be the event that the ball is drawn from bag “A” and E2 be the... View Article
One Die Is Thrown Three Times And The Sum Of The Thrown Numbers Is 15 The Probability For Which Number 4 Appears In First Throw 1) 1/18 2) 1/36 3) 1/9 4) 1/3 Solution: Option (1) 1/18 Let A be the event of getting a sum of 15 of the three thrown numbers and B... View Article
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