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Quantitative Aptitude

PROBABILITY MCQs

Probability, Probability I

Total Questions : 775 | Page 11 of 78 pages
Question 101. Two fair dice  are rolled simultaneously.  It is found that one of the dice show odd prime number.  The probability that the remaining dice also show an odd prime number, is equal to
  1.    15
  2.    25
  3.    35
  4.    45
 Discuss Question
Answer: Option A. -> 15
:
A
Odd prime on a die are 3 and 5.
Let event A = one of the dice show odd prime number, and event B = remaining dice also shows odd prime number.
P(BA)=n(AB)n(A).A={(3,1)(3,2),(3,3),(3,4),(3,5),(3,6),(5,1),...(5,6),(1,3),(2,3),...(6,3),(1,5),...(5,5),}AndAB={(3,3).(5,5),(3,5)(5,3)}P(AB)=420=15.
Question 102. On a toss of two dice, A throws a total of 5.  Then the probability that he will throw another 5 before he throws 7 is
  1.    245
  2.    25
  3.    181
  4.    19
 Discuss Question
Answer: Option B. -> 25
:
B
let P(A) be the probability of throwing total of 5 and P(B) the probability of throwing total 7.
P(A)=436=19,P(B)=636=16,P(AorB)=518andP(AandB)=0P(neitherAnorB)=1318,P(5before7)=19+1318.19+1318.1318.19+....=19[111318]=25.
Question 103. A coin whose faces marked 2 and 3 is thrown 5 times, then chance of obtaining a total of 12 is
  1.    516
  2.    58
  3.    532
  4.    524
 Discuss Question
Answer: Option A. -> 516
:
A
It is given that the coin has faces 2 and 3, tossed five times.
To get sum 12 we need 2,2,2,3,3 in any combination.
So, in the 5 throws, there need to be 3 2's. Automatically other 2 will be 3's since there is no other possibility.
So, it boils down to choosing 3 from 5.
Therefore P(getting sum 12) =5C3(12)5=516.
Question 104. The probability that a certain beginner at golf gets good shot if he uses correct club is 13, and the probability of a good shot with an incorrect club is 14.  In his bag there are 5 different clubs only one of which is correct for the good shot.  If he chooses a club at random and take a stroke, the probability that he gets a good shot is
  1.    13
  2.    112
  3.    415
  4.    712
 Discuss Question
Answer: Option C. -> 415
:
C
P( getting correct club ) =15
Therefore P(hitting good shot by correct club) =13×15=115
P(getting wrong club) =45
P(hitting good shot with wrong club) =45×14=15.
Therefore P(hitting good shot) =115+15=415.
Question 105. A bag contains 'a' white and 'b' black balls. Two players A  and B alternately draw a ball from the bag, replacing the ball each time after the draw.  A begins the game.  If the probability of A winning ( that  is
drawing a white ball) is twice the probability of B winning, then the ratio a : b  is equal to
  1.    1 : 2
  2.    2 : 1
  3.    1 : 1
  4.    1 : 3
 Discuss Question
Answer: Option C. -> 1 : 1
:
C
Let the event when a white ball is draw be W and for black ball let it be B. So A wins when we get sequence of the form
W or WBW or WBBBW or WBBBBBW......
Probability of getting W is a/a+b . So we get
P(A)=aa+b+(ba+b)2aa+b+(ba+b)4aa+b+...=a+ba+2bSimilarlywegetP(B)=ba+b.aa+b+(ba+b)3aa+b+...=ba+2bP(A)=2P(B)a+b=2ba=b.
Question 106. A bag contains 3 white, 3 black and 2 red balls.  One by one, three balls are drawn without replacing them.  Then the probability that the third ball is red , is given by
  1.    524
  2.    112
  3.    14
  4.    12
 Discuss Question
Answer: Option C. -> 14
:
C
Number of white balls = 3, black balls = 3 red balls = 2
Since drawn balls are not replaced, for third ball to red, we have following patterns:
E1=WWR,BBR,E2=BWR,WBR;andE3=RBR,RWR,BRR,WRR
P(E1)=2×38×27×26,P(E2)=2×38×37×26,P(E3)=4×2×3×18.7.6
Required probability =P(E1)+P(E2)+P(E3)=14
Question 107. A natural number   is chosen at random from the first 100 natural numbers. Then the probability, for the in-equation x+100x>50 satisfied, is
  1.    120
  2.    1120
  3.    13
  4.    14
 Discuss Question
Answer: Option B. -> 1120
:
B
x+100x>50x250x+100>01x<25521,or25+521<x100(1x100,xϵN)
x=1,2,48,49,50,...,100
Therefore the total number of ways = 100 and favorable ways = 55
Required probability =55100=1120.
Question 108. In a multiple choice question there are four alternative answers of which one or more than one  is or are correct.  A candidate will get marks on the question only if he ticks all correct answers.  The candidate decides to tick answers at random.  If he is allowed up to three chances to answer the question, the probability that he will get marks on it is given by
  1.    1−(1415)3
  2.    (115)3
  3.    15
  4.    1415
 Discuss Question
Answer: Option C. -> 15
:
C
The total number of ways of answering one or more alternatives of 4 is 4C1+4C2+4C3+4C4=15,
Out of these 15 combinations, only one combination is correct. The probability of answeringthe alternaive correctly at the first trial is 115, that at the second trial is (1415.114)=115, and that at the third trail is 1415.1314.113=115,
Therefore the probability that the andidate will get marks on the question if he allowed upto three chances, is 115+115+115=15.
Question 109. The odds in favour of A solving a problem are 3 to 4 and the odds  against B solving the same problem are 5 to 7.  If they  both try the problem, the probability that the problem is solved is:
  1.    4184
  2.    1621
  3.    521
  4.    14
 Discuss Question
Answer: Option B. -> 1621
:
B
P(A)=37,P(B)=712P(¯A)=47,P(¯B)=512P(AB)=1P(¯¯¯¯¯¯¯¯¯¯¯¯¯¯AB)=1P(¯A¯B)P(AB)=147×512=1621
Question 110. Urn A contains 6 red , 4 white balls and urn B contains 4 red and 6 white balls.  One ball is drawn from the urn A and placed in the urn B.  Then one ball is drawn at random from urn B and placed in the urn A.  if one ball is now drawn from the urn A, then the probability that it is found to be red is
  1.    3255
  2.    3040
  3.    3250
  4.    34
 Discuss Question
Answer: Option A. -> 3255
:
A
case(i): red ball from A to B, red ball from B to A ,then red ball from A
P1=610×511×610
Case(ii): red ball from A to B, white ball from B to A, red ball from A
P2=610×611×510
Case(iii): white ball from A to B, red ball from B to A, red ball from A
P3=410×411×710
Case(iv): white ball from A to B, white ball from B to A, red ball from A
P4=410×711×610
Therefore required probability =P1+P2+P3+P4=6401100=3255

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