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Probability Quiz-7

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. If (1+4 p)/p,(1-p)/4,(1-2 p)/2 are probabilities of three mutually exclusive events, then
  •  1/3 ≤ p ≤ 1/2
  •  1/2 ≤ p ≤ 2/3
  •  1/6 ≤ p ≤ 1/2
  •  None of these
Solution
(d) Since (1+4 p)/p,(1-p)/2 and (1-2 p)/2 are probabilities of three mutually exclusive events 
 ∴0 ≤ (1+4 p)/p 1,0 ≤ (1-p)/2 ≤ 1,0 ≤ (1-2 p)/2 ≤ 1 and,
 0 ≤ (1+4 p)/p+(1-p)/2+(1-2 p)/2 ≤ 1 
 ⇒-1/4 ≤ p ≤ 3/4,-1 ≤ p ≤ 1,-1/2 ≤ p ≤ 1/2 and 1/2 ≤ p ≤5/2
 ⇒max⁡{-1/4,-1,-1/2,1/2}≤ p ≤ min⁡{3/4,1,1/2,5/2} 
 ⇒1/2 ≤ p ≤ 1/2 
⇒ p = 1/2

Q2.If m rupee coins and n ten paise coins are placed in a line, then the probability that the extreme coins are ten paise coins, is
  •  m+nCm
  •  (n(n-1))/((m+n)(m+n-1))
  •  m+nPm
  •  m+nPn
Solution
(b) m one rupee coins and n ten paise coins can be placed in a line in ((m+n) !)/(m !n !) ways 
 ∴ Total number of elementary events =((m+n) !)/(m !n !) 
 If the extreme coins are ten paise coins, then the remaining (n-2) ten paise coins and m one rupee coins can be arranged in a line a ((m+n-2) !)/(m !(n-2) !) ways 
 ∴ Favourable number of elementary events =((m+n-2) !)/(m !(n-2) !) 
 ∴ Required probability =(((m+n-2) !)/(m !(n-2) !))/(((m+n) !)/(m !n !))=(n(n-1))/((m+n)(m+n-1))
Q3.  A bag contains 10 white and 3 black balls. Balls are drawn one-by-one without replacement till all the black balls are drawn. The probability that the procedure of drawing balls will come to an end at the seventh draw is
  •   105/286
  •  15/286
  •  181/286
  •  None of these
Solution
(b) Consider the following events : 
 A= Getting 2 black balls and 4 white in first 6 draws 
 B= Getting a black ball in 7th draw 
 Required probability =P(A∩B)=P(A)P(B/A) 
 ⇒Required probability=( 3 C2× 10 C4)/( 13 C6 )×1/7=15/286

Q4. A sample of a 4 times is drawn at a random without replacement from a lot of 10 items containing 3 defectives. If x denotes the number of defective items in the sample, then P(0&#60x&#603)is equal to
  •  3/10
  •  4/5
  •  1/2
  •  1/6
Solution

 


Q5.An urn contains 4 white and 3 red balls. There balls are drawn with replacement from this urn. Then, the standard deviation of the number of red balls drawn is
  •  6/7
  •  36/49
  •  5/7
  •  25/49
Solution
(a) Let X denotes the number of red balls. 
Here probability of getting red balls, p=3/7 and probability of getting red bills, 
q=4/7 
 P1 (X=0)=3C0 (3/7)0 (4/7)3=64/(7)3
P2 (X=1)=3C1 (3/7)1 (4/7)2=144/(7)3 
 P3 (X=2)=3C2 (3/7)2 (4/7)1=108/(7)3 

 


Q6. A and B appeared for an interview for two posts. The probability of A's selection is 1/3 and that of B's selection is 2/5. The probability that only one of them is selected, is
  •  7/15
  •  8/15
  • 2/15
  •  4/15
Solution
(a) Explaination not available

Q7.If P(B)=3/4,P(A∩B∩Cc)=1/3 and P(Ac∩B∩Cc)=1/3,then P(B∩C) is
  •  1/12
  •  1/6
  •  1/9
  •  1/15
Solution

 


Q8.A box contains 3 white and 2 red balls. If we draw one ball and without replacing the first ball, the probability of drawing red ball in the second draw is
  •  8/25
  •  2/5
  •  3/5
  •  21/25
Solution
(b) There are two cases arise. 
 Case I 
If Ist ball is white, then P=3C1/5C1 ×2C1/3C1 =6/20=3/10 
 Case II
 If Ist ball is red, then P=2C1/5C1 ×1C1/4C1 =2/20=1/10 
 ∴ Required probability =3/10+1/10=2/5

Q9.If a coin be tossed n times, then probability that the head comes odd times is
  •  1/2
  •  1/2n
  •  1/2n-1
  •  None of these
Solution
(a) A coin is tossed n times. 
 ∴ Total number of ways =2n 
 If head comes odd times, then favourable ways =2n-1
 ∴ Required probability of getting odd times head =2n-1/2n =1/2

Q10. Suppose n(≥3) persons are sitting in a row. Two of them are selected at random. The probability that they are not together is
  •  1-2/n
  •  2/(n-1)
  •  1-1/n
  • None of these
Solution
(a) The total number of ways of choosing 2 persons out of n is nC2 
 After selecting two persons when the remaining (n-2) persons sit in row (n-1) places are created between them in which 2 persons can be arranged in n-1C2×2 ! ways 
 So, required probability =( nC2×2 !)/( nC2 )=(n-2)/n=1-2/n

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