Theorems of probability

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Two different numbers are taken from the set {0, I ,2,3,4,5,6,7,8, 9, 10).The probability that their sum and positive difference, are both multiple of 4, is $\frac{x}{55}$ then x equals

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Solution

$Let the two numbers are ' a' and ' b'$
$According to the question a + b = 4 p and a - b = 4 q where p, q \in I$
$\begin{array}{l} \Rightarrow 2 a = 4 (p + q) and 2 b = 4 (p - q) \\ \Rightarrow a = 2 I_{1} and b = 2 I_{2} \end{array}$
Hence both a and b must be even.
$Hence n (S) =^{11} C_{2}$
$\begin{array}{l} n (A) = (0, 4), (0, 8), (2, 6), (2, 10), (4, 8), (6, 10) = 6 \\ ∴ P (A) = \frac{6}{^{11} C_{2}} = \frac{6}{55} \end{array}$

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Theorems of probability

Remember concepts with our Masterclasses.

Two different numbers are taken from the set {0, I ,2,3,4,5,6,7,8, 9, 10).The probability that their sum and positive difference, are both multiple of 4, is x55 then x equals

Let the two numbers are ' a ' and ' b ' According to the question a+b=4p and a−b=4q where p, q∈I⇒ 2a=4(p+q) and 2b=4(p−q)⇒ a=2I1 and b=2I2Hence both a and b must be even. Hence n(S)=11C2 n(A)=(0,4),(0,8),(2,6),(2,10),(4,8),(6,10)=6∴ P(A)=6 11C2=655

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Two different numbers are taken from the set {0, I ,2,3,4,5,6,7,8, 9, 10).The probability that their sum and positive difference, are both multiple of 4, is $\frac{x}{55}$ then x equals