Question
In a class, 30% of the students offered English, 20% offered Hindi and 10% offered both. If a student is selected at random, What is the probability that he has offered English or Hindi ?
Answer: Option A
$$\eqalign{
& P(E) = \frac{{30}}{{100}} = \frac{3}{{10}}; \cr
& P(H) = \frac{{20}}{{100}} = \frac{1}{5}\,\,{\text{and}} \cr
& {\text{P }}\left( {{\text{E}} \cap {\text{H}}} \right)\,{\text{ = }}\frac{{10}}{{100}} = \frac{1}{{10}} \cr} $$
P (E or H) = $$P(E \cup H)$$
= P(E) + P(H) - $$P(E \cup H)$$
$$\eqalign{
& = \left( {\frac{3}{{10}} + \frac{1}{5} - \frac{1}{{10}}} \right) \cr
& = \frac{4}{{10}} \cr
& = \frac{2}{5} \cr} $$
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$$\eqalign{
& P(E) = \frac{{30}}{{100}} = \frac{3}{{10}}; \cr
& P(H) = \frac{{20}}{{100}} = \frac{1}{5}\,\,{\text{and}} \cr
& {\text{P }}\left( {{\text{E}} \cap {\text{H}}} \right)\,{\text{ = }}\frac{{10}}{{100}} = \frac{1}{{10}} \cr} $$
P (E or H) = $$P(E \cup H)$$
= P(E) + P(H) - $$P(E \cup H)$$
$$\eqalign{
& = \left( {\frac{3}{{10}} + \frac{1}{5} - \frac{1}{{10}}} \right) \cr
& = \frac{4}{{10}} \cr
& = \frac{2}{5} \cr} $$
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