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The letter of the word “PROBABILITY” are written down at random in a row. Let $E_{1}$ denotes the event that two I’s are together and $E_{2}$ denotes the event that two B’s are together, then
Column-I
Column-II
A) $P (E_{1}) = P (E_{2})$ P) $\frac{2}{55}$
B) $P (E_{1} \cap E_{2})$ Q)    $\frac{2}{11}$
C) $P (E_{1} \cup E_{2})$ R)    $\frac{1}{5}$
D) $P (\frac{E_{1}}{E_{2}})$ S)    $\frac{18}{55}$

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$A \to Q; B \to P; C \to S; D \to R$

$A \to P; B \to Q; C \to S; D \to R$

$A \to P; B \to Q; C \to R; D \to S$

$A \to Q; B \to S; C \to P; D \to R$

answer is A.

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Detailed Solution

$P; R; O; B; A; B; I; L; I; T; Y$

$n (s) = \frac{11!}{2! 2!} P (E_{1}) = P (E_{2}) = \frac{n (E)}{n (S)} = \frac{10!}{2!} \times \frac{2! 2!}{11!} = \frac{2}{11} P (E_{1} \cap E_{2}) = P (E_{1}) + P (E_{2}) - P (E_{1} \cap E_{2}) [P; R; O; A; L; T; Y; B B; I I] ∴ \frac{9!}{(\frac{11!}{2! 2!})} = \frac{2}{55} P (E_{1} \cup E_{2}) = P (E_{1}) + P (E_{2}) - P (E_{1} \cap E_{2}) P (\frac{E_{1}}{E_{2}}) = \frac{P (E_{1} \cap E_{2})}{P (E_{2})}$

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