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RRB Technician Grade I 13 March 2026 Paper

Question : 47 of 100

Marks: +1, -0

P(E)=\frac{1}{6}

P(F)=\frac{1}{2}

P(not E and not F)

Solution:

Concept:

For mutually exclusive events,

P(E \cap F) = 0

, and

P(\text{not E and not F})

is the complement of the union

P(E \cup F)

Explanation:

Since E and F are mutually exclusive, they cannot occur together, so

P(E \cap F) = 0

Using the addition theorem:

P(E \cup F) = P(E) + P(F) - P(E \cap F)

Substitute the given values:

P(E \cup F) = \frac{1}{6} + \frac{1}{2} - 0

Convert to like fractions:

P(E \cup F) = \frac{1}{6} + \frac{3}{6} = \frac{4}{6} = \frac{2}{3}

Now,

P(\text{not E and not F}) = P(E' \cap F')

By De Morgan's law,

P(E' \cap F') = 1 - P(E \cup F)

Therefore,

P(\text{not E and not F}) = 1 - \frac{2}{3} = \frac{1}{3}

Answer:

The required probability is

\frac{1}{3}

, which matches option D.

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