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As we know that the general   term   in the binomial expansion of    is given by  So,  Term in  the expansion of  :  Term in  the expansion of  :  Term in  the expansion of  : Now, As given in the question, From here, we get , Which can be written as  From these equations we get,   
As we know that if a point moves in a plane in such a way that its distance from two-point remain constant then the path is an ellipse. Now, According to the question, the distance between the point from where the sum of  the distance from a point is constant =  10 Now, the distance between the foci=8 Now, As we know the relation, Hence the equation of the ellipse is, Hence the path of...
It is given that A = {–1, 1} A is an non-empty set  Therefore, Lets first find  Now,
Total numbers of numbers in the draw = 20 Numbers to be selected = 6  n(S) =   Let E be the event that six numbers match with the six numbers fixed by the lottery committee. n(E) = 1 (Since only one prize to be won.)  Probability of winning =
Sample space when three coins are tossed: [Same as a coin tossed thrice!] S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT} Number of possible outcomes, n(S) = 8                             [Note: 2x2x2 = 8] Let E be the event of getting exactly 2 tails = {TTH, HTT, THT}  n(E) = 3      The required probability of getting exactly 2 tails is .
Given problem is Now, we will reduce it into                                                                                                                                                                                                                                         Now, multiply numerator an denominator by                                            Therefore, answer is   
Given, Each wheel can be labelled with 10 digits.  Number of ways of selecting 4 different digits out of the 10 digits =  These 4 digits can arranged among themselves is  ways.  Number of four digit numbers without repetitions = Number of combination that can open the suitcase = 1  Required probability of getting the right sequence to open the suitcase =  
(ii) Since 4-digit numbers greater than 5000 are to be formed, The  place digit can be filled up by either 7 or 5 in  ways Since repetition is not allowed,  The remaining 3 places can be filled by remaining 4 digits in   ways.   Total number of 4-digit numbers greater than 5000 = We know, a number is divisible by 5 if unit’s place digit is either 0 or 5. Case 1. When digit at  place is 5, the...
(i) Since 4-digit numbers greater than 5000 are to be formed, The  place digit can be filled up by either 7 or 5 in  ways Since repetition is allowed,  Each of the remaining 3 places can be filled by any of the digits 0, 1, 3, 5, or 7 in  ways.   Total number of 4-digit numbers greater than 5000 =  (5000 cannot be counted, hence one less) We know, a number is divisible by 5 if unit’s place...
Given, Total number of persons = 5 No. of male spokesperson  = 3 No. of spokesperson who is over 35 years of age = 2 Let E be the event that the spokesperson is a male and F be the event that the spokesperson is over 35 years of age.    Since only one male is over 35 years of age,   We know, Therefore, the probability that the spokesperson will either be a male or over 35 years of age is .
Given, P(A) = 0.54, P(B) = 0.69, P(A  B) = 0.35 And, P(A  B) = 0.88 (iv) P(B  A') = P(B-A) = P(B) - P(A  B)  = 0.69 - 0.35 = 0.34  P(B  A') = 0.34
Given, P(A) = 0.54, P(B) = 0.69, P(A  B) = 0.35 And, P(A  B) = 0.88 (iii) P(A  B') = P(A-B) = P(A) - P(A  B) = 0.54 - 0.35 = 0.19   P(A  B') = 0.19
Given, P(A) = 0.54, P(B) = 0.69, P(A  B) = 0.35 (i) We know, P(A   B) = P(A) + P(B) - P(A  B)  P(A   B) = 0.54 + 0.69 - 0.35 = 0.88  P(A   B) = 0.88
Given, 3 letters are put in 3 envelopes. The number of ways of putting the 3 different letters randomly = 3! Number of ways that at least one of the 3 letters is in the correct envelope   = No. of ways that exactly 1 letter is in correct envelope +  No. of ways that 2 letters are in the correct envelope(The third is automatically placed correctly) = No. of ways that exactly 1 letter is in...
Total number of students = 100 Let A and B be the two sections consisting of 40 and 60 students respectively. We found out in (a) that the probability that both students are in same section is  (b) Probability that both the students are in different section =   
Total number of students = 100 Let A and B be the two sections consisting of 40 and 60 students respectively. Number of ways of selecting 2 students out of 100 students.=  (a)  If both are in section A: Number of ways of selecting 40 students out of 100 =   (The remaining 60 will automatically be in section B!) Remaining 38 students are to be chosen out of (100-2 =) 98 students  Required...
Given, Total number of tickets sold = 10,000 Number of prizes awarded = 10 (c) If ten tickets are bought: Number of tickets not awarded = 10000 - 10 = 9990  P(not getting a prize) =   
Given, Total number of tickets sold = 10,000 Number of prizes awarded = 10 (b) If two tickets are bought: Number of tickets not awarded = 10000 - 10 = 9990  P(not getting a prize) = 
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