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in Introduction to Probability Theory by (47.8k points)
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If a and b are chosen randomly from the set {1, 2, 3, 4} with replacement, then the probability of the real roots of the equation x2 + ax + b = 0 is

(a) 3/16

(b) 5/16

(c) 7/16

(d) 11/16

1 Answer

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by (47.4k points)
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Best answer

(c) 7/16

x2 + ax + b = 0 ⇒ \(x = {-a \pm \sqrt{a^2-4b} \over 2}\)

Given that the roots are real ⇒ a2 – 4b ≥ 0 or a2 > 4b 

When a = 1, b = 1 or 2 or 3 or 4, a2 – 4b < 0 

When a = 2, b = 1, a2 – 4b = 0 

When a = 3, b = 1 or 2 for which a2 – 4b ≥ 0 

When a = 4, b = 1 or 2, 3 or 4 for which a2 – 4b ≥ 0 

So, Selecting from the 4 number 42 = 16 ways. 

(i.e.,) n(s) = 16 

n(A) = (2 or 3 or 4) = 3 

n(B) = (1 or 2 or 3 or 4) = 4 

P(A) + P(B) = 3/16 + 4/16 = 7/16

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