Question
There are 8 boys and 4 girls giving interview for a
job. If three of them are selected, then what is the probability that one of the three is a girl and the other two are the boys?Solution
Total Number of candidates = 12 Let ‘S’ be the Sample Space Then, n(S) = number of ways of three candidates who got selected n(S) = 12C3 = 220 Let ‘E’ be the event of 1 girl and 2 boys selected Therefore, n(E) = number of possibility of 1 girl out of 4 and 2 boys out of 8 n(E) = 4C1 × 8C2 = 4 × 28 = 112 Now the required Probability = 112/220 = 28/55
I). 5p2 Â - p - 4 = 0
II). q2 - 12q + 27 = 0
I. y/16 = 4/yÂ
II. x3 = (2 ÷ 50) × (2500 ÷ 50) × 42 × (192 ÷ 12)
I). p2 + 8p + 15 = 0
II). q2 + 9q + 20 = 0
I. 2x² + 11x + 12 = 0
II. 2y² + 19y + 45 = 0
I:Â x2Â - 33x + 242 = 0
II:Â y2Â - 4y - 77 = 0
I. 6x² - 23x + 7 = 0
II. 6y² - 29y + 9 = 0
I. 6x2 + 23x + 10 = 0
II. 2y2 - 3y - 5 = 0
I. 8/(21x) - 2/7 = 0
II. 16y² - 24y +9 = 0
In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between 'p' and 'q' and choose...
Solve the quadratic equations and determine the relation between x and y:
Equation 1: 4x² - 12x + 9 = 0
Equation 2: 2y² + 8y + 6 = 0
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