An integer is chosen at random from the first two hundred positive integers. What is the probability that the integer chosen is divisible by 6 or 8 ? -Maths 9th

An integer is chosen at random from the first two hundred positive integers. What is the probability that the integer chosen is divisible by 6 or 8 ? -Maths 9th
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Answer :

As there are 200 integers, total number of exhaustive, mutually exclusive and equally likely cases, i.e, n(S) = 200 Let A : Event of integer chosen from 1 to 200 being divisible by 6⇒ n(A) = 33 \(\bigg(rac{200}{6}=33rac{1}{3}\bigg)\)Let B : Event of integer chosen from 1 to 200 being divisible by 8⇒ n(B) = \(rac{200}{8}\) = 25A ∩ B : Event of integer chosen from 1 to 200 being divisible by both 6 and 8, i.e, divisible by 24 (∵ LCM of 6 and 8 = 24)⇒ n(A ∩ B) = 8 \(\bigg(rac{200}{24}=8( ext{approx})\bigg)\)∴ P(A ∪ B) = P(Integer chosen is divisible by 6 or 8)= P(A) + P(B) – P(A ∩ B) = \(rac{n(A)}{n(S)}\) + \(rac{n(B)}{n(S)}\) - \(rac{n(A\cap{B})}{n(S)}\)= \(rac{33}{200}\) + \(rac{25}{200}\) - \(rac{8}{200}\) = \(rac{50}{200}\) = \(rac{1}{4}\).
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Description : If the polynomial x^6 + px^5 + qx^4 – x^2 – x – 3 is divisible by x^4 – 1, then the value of p^2 + q^2 is : -Maths 9th

Last Answer : The divisor is x4−1=(x−1)(x+1)(x2+1) By factor theorem, f(1)=f(−1)=0 Thus, 1+p+q−1−1−3=0 and 1+q−1−3=p−1 i.e., p+q=4 and p−q=−2 Adding the two, 2p=2 i.e. p=1 and ∴ q=3. ∴ p2+q2=1+9=10

The product of (q-1) consecutive integers where `qgt1` is divisible by ___________

Description : The product of (q-1) consecutive integers where `qgt1` is divisible by ___________

Last Answer : The product of (q-1) consecutive integers where `qgt1` is divisible by ___________