Question

In a high school having equal number of boy students and girl students, 75% of the students study Science and the remaining 25% students study Commerce. Commerce students are two times more likely to be a boy than are Science students. The amount of information gained in knowing that a randomly selected girl student studies Commerce (rounded off to three decimal places) is _______ bits.

Answer 1

Concept:

Information (I):

\(I = {\log _2}\frac{1}{P}\) bits

Calculation:

∴ No. of boys and girl students are equal

∴ \(P\left( B \right) = \frac{1}{2}\;\;\& \;\;P\left( G \right) = \frac{1}{2}\)

∵ 75% students study science and 25% students study commerce.

∴ \(P\left( S \right) = \frac{3}{4}\;\;\& \;\;P\left( C \right) = \frac{1}{4}\)  

∵ commerce students are two times more likely to be a boy than are science students.

\(P\left( {\frac{B}{C}} \right) = 2P\left( {\frac{B}{S}} \right)\)

Now;

\(P\left( B \right) = \left( {\frac{B}{C}} \right) \cdot P\left( C \right) + P\left( {\frac{B}{S}} \right) \cdot P\left( S \right)\)

\(\frac{1}{2} = P\left( {\frac{B}{C}} \right) \cdot \frac{1}{4} + \frac{1}{2}P\left( {\frac{B}{C}} \right) \cdot \frac{3}{4}\)

\(P\left( {\frac{B}{C}} \right) = \frac{4}{5}\)

Now,

\(P\left( {\frac{C}{B}} \right) = \frac{{P\left( {\frac{B}{C}} \right) \cdot P\left( C \right)}}{{P\left( B \right)}} = \frac{{\frac{4}{5} \times \frac{1}{4}}}{{\frac{1}{2}}} = \frac{2}{5}\)

\(P\left( C \right) = P\left( {\frac{C}{B}} \right)P\left( B \right) + P\left( {\frac{C}{G}} \right) \cdot P\left( G \right)\)

\(\frac{1}{G} = \frac{2}{5} \times + P\left( {\frac{C}{G}} \right)\frac{1}{2}\)

\(P\left( {\frac{C}{G}} \right) = \frac{1}{{10}}\)

The amount of information gained in knowing that a randomly selected girl student studies commerce

\(I = \log_2[ \frac{1}{{P\left[ {\frac{C}{G}} \right]}} ]= - {\log _2}P\left[ {\frac{C}{G}} \right]\)

= log₂10

= 3.333 bits.