A number $x_{1} x_{2} x_{3} x_{4}$ is selected at random from the set of all 4-digit natural numbers, $x_{1}, x_{2}, x_{3}, x_{4}$ being the digits of the number.
Let A be the event that $x_{1} = x_{2},$ B be the event that $x_{1} < x_{3},$ C be the event $x_{2} = x_{3},$
D be the event that $x_{2} < x_{3}$
If $P (E)$ denotes the probability of event $E,$ then

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$P (\frac{B}{A}) = \frac{2}{5}$

$P (\frac{A}{C}) = \frac{1}{10}$

$P (\frac{B}{D}) = \frac{4}{9}$

$P (\frac{A}{B}) = \frac{1}{10}$

answer is A, B, D.

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Detailed Solution

$x_{1}, x_{2}, x_{3}, x_{4} \in {0, 1, 2, ...., 9} but x_{1} \neq 0 n (s) = 9 \times 10^{3} = 9000 x_{1} = x_{2} \Rightarrow no. of favorable cases to A = 9 \times 1 \times 10 \times 10 x_{1} < x_{3} \Rightarrow no. of favorable cases to B = (^{9} C_{2}) \times 10 \times 10 x_{2} = x_{3} \Rightarrow no. of favorable cases to C = (^{10} C_{1}) \times 9 \times 10 x_{2} < x_{3} \Rightarrow no. of favorable cases to D = (^{10} C_{2}) \times 9 \times 10 ∴ P (A) = \frac{1}{10}, P (B) = \frac{36}{90} = \frac{2}{5}, P (C) = \frac{1}{10}, P (D) = \frac{45}{100} = \frac{9}{20}$