Quantitative Aptitude
PROBABILITY MCQs
Probability, Probability I
Total Questions : 775
| Page 11 of 78 pages
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(A⋂B)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),}AndA⋂B={(3,3).(5,5),(3,5)(5,3)}∴P(AB)=420=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(A⋂B)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),}AndA⋂B={(3,3).(5,5),(3,5)(5,3)}∴P(AB)=420=15.
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[11−1318]=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[11−1318]=25.
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.
:
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
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.
:
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
drawing a white ball) is twice the probability of B winning, then the ratio a : b is equal to
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=2b⇒a=b.
:
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=2b⇒a=b.
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
:
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
Answer: Option B. -> 1120
:
B
x+100x>50⇒x2−50x+100>0⇒1≤x<25−5√21,or25+5√21<x≤100(1≤x≤100,xϵN)
⇒x=1,2,48,49,50,...,100
Therefore the total number of ways = 100 and favorable ways = 55
Required probability =55100=1120.
:
B
x+100x>50⇒x2−50x+100>0⇒1≤x<25−5√21,or25+5√21<x≤100(1≤x≤100,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
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.
:
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.
Answer: Option B. -> 1621
:
B
P(A)=37,P(B)=712P(¯A)=47,P(¯B)=512P(A∪B)=1−P(¯¯¯¯¯¯¯¯¯¯¯¯¯¯A∪B)=1−P(¯A∩¯B)∴P(A∪B)=1−47×512=1621
:
B
P(A)=37,P(B)=712P(¯A)=47,P(¯B)=512P(A∪B)=1−P(¯¯¯¯¯¯¯¯¯¯¯¯¯¯A∪B)=1−P(¯A∩¯B)∴P(A∪B)=1−47×512=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
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
:
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