




Step 1: The reaction involves the oxidation of toluene with \(\text{CrO}_2\text{Cl}_2\) and \(\text{CS}_2\). \(\text{CrO}_2\text{Cl}_2\) is a strong oxidizing agent, typically oxidizing the methyl group (\(\text{-CH}_3\)) in toluene to a carboxylic acid (\(\text{COOH}\)) group. This results in a formyl group (\(\text{CHO}\)) being left behind in residue (A), as toluene is partially oxidized.
Step 2: The treatment with water and NaHSO\(_3\) ensures further oxidation of the formyl group (\(\text{CHO}\)) into a carboxyl group (\(\text{COOH}\)). Diluting with HCl gives a sodium salt of the carboxyl group (\(\text{COONa}\)) in compound (B).
Thus, the structures of residue (A) and compound (B) are as follows: Residue (A) has a formyl group (\(\text{CHO}\)) and compound (B) has a carboxylate group (\(\text{COONa}\)).
Observe the following reactions (unbalanced):
\( P_2O_3 + H_2O \rightarrow X \)
\( P_4O_{10} + H_2O \rightarrow Y \)
The number of \( P=O \) bonds present in \( X, Y \) are respectively:
If a random variable \( x \) has the probability distribution 
then \( P(3<x \leq 6) \) is equal to