Probability Quiz-8

Dear Readers,

As per analysis for previous years, it has been observed that students preparing for JEE MAINS find Mathematics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Mathematics due to their ENGINEERING background. 

Furthermore, sections such as Mathematics are dominantly based on theories, laws, numerical in comparison to a section of Engineering which is more of fact-based, Physics, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced. 

Q1. Three dice are thrown simultaneously, then probability of throwing a total greater than 4 is
  •  1/54
  •  53/54
  •  5/108
  •  None of these
(b) P(getting a sum greater than 4) =1-P(getting a sum less than 5) …(i) 
 For sum 3, Number of cases=1 
 For sum 4, Number of cases=3 
 ∴ Total number of cases=4 
 From Eq.(i), 

Q2.An integer is chosen at random from first two hundred numbers. Then, the probability that the integer chosen is divisible by 6 or 8 is
  •  1/4
  •  2/4
  •  3/4
  •  None of these
(a) One integer can be chosen out of 200 integers in 200 C1ways. 
Let A be the event that an integer selected is divisible by 6 and B that it is divisible by 8 Then,P(A)=33/200,P(B)=25/200 and P(A∩B)=8/200 
 ∴P(A∪B)=P(A)+P(B)-P(A∩B) =33/200+25/200-8/200=1/4
Q3.  Given two mutually exclusive events A and B such that P(A)=0.45 and P(B)=0.35,P(A∩B) is equal to
  •   63/400
  •  0.8
  •  63/200
  •  0

Q4. Suppose f(x)=x3+ax2+bx+c,where a,b,c are chosen respectively by throwing a dice three times. Then, the probability that f(x) is an increasing function, is
  •  4/9
  •  3/8
  •  2/5
  •  16/34
(a) ∵f(x)=x3+ax2+bx+c 
 ∴f' (x)=3x2+2ax+b y=f(x) is increasing. 
 ⇒f' (x)≥0,∀x And for f'(x)=0 should not form an interval. 
 This is true for exactly 16 ordered pairs (a,b),1≤a,b≤6 namely (1, 1), (1, 2), (1, 3), (1, 4); (1, 5), (1, 6), (2, 2),(2, 3), (2, 4), (2, 5), (2, 6); (3, 3),(3, 4),(3, 5),(3, 6) and (4, 6). 
 Thus, required probability =16/36=4/9

Q5.If A and B are any two events, then P(Ac∪B) is equal to
  •  P(Ac)P(Bc)
  •  1-P(A)-P(B)
  •  P(A)+P(B)-P(A∩B)
  •  P(B)-P(A∩B)


Q6. If P(A∩B)=1/2,P(Ac∩Bc )=1/3,P(A)=p,P(B)=2 p, then the value of p is given by
  •  1/3
  •  7/18
  • 4/9
  •  1/2
(b) We have, P(Ac∩Bc )=1/3 
⇒p+2 p-1/2=2/3

Q7.One card is drawn randomly from a pack of 52 cards, then the probability that it is a king of spade, is
  •  1/26
  •  3/26
  •  4/13
  •  3/13
(c) Probability of getting a king =4/52=1/13 
 Probability of getting a spade =13/52=1/4
 Probability of getting king and a spade =1/52 
 ∴P(king or spade)=P(king)+P(spade)-P(king and spade) 

Q8.The probability that the three cards drawn from a pack of 52 cards, are all black, is
  •  1/17
  •  2/17
  •  3/17
  •  2/19
(b) In a pack of 52 cards, there are 26 black cards. 
 ∴Required probability=( 26 C3)/( 52 C3 ) =(26×25×24)/(3×2×1)×(3×2×1)/(52×51×50) =2/17

Q9.A biased coin with probability p,0&#60p&#601 of heads is tossed until a head appears for the first time. If the probability that the number of tossed required is even is 2/5,then p equals
  •  1/3
  •  2/3
  •  2/5
  •  3/5
(a) Let q=1-p. 
Since, head appears first time in an even throw 2 or 4 or 6 
 ∴ 2/5=qp+q3p+q5p+... 
 ∴ 2/5=qp/(1-q2 )
 ⇒ 2/5=(1-p)p/(1-(1-p)2
 ⇒ 2/5=(1-p)/(2-p) 
 ⇒ 4-2P=5-5P 
⇒ p=1/3

Q10. A matrix is chosen at random from the set of all 2×2 matrices with elements 0 and 1 only. The probability that the value of the determinant of the matrix chosen is positive, is
  •  1/2
  •  3/16
  •  11/16
  • 13/16

 Hence, the required probability =3/16

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