A bag contained 3 maths book and 2 physics books. A book is drawn at random if it is of math, 2 more books of maths together with this book put back in the bag and if it is of physics it is not replaced in the bag. This experiment is repeated 3 time. If third draw gives math book, The probability that first two drawn books were of physics is $\frac{p}{q}$ (where $H.C.F(p, q) = 1$) then $q - 8(p+1) = \ldots\ldots\ldots\ldots$
Step-by-Step Solution
Key Concept: Use Bayes' theorem to find P(both physics | third is math) by calculating likelihoods for all possible outcomes of first two draws that lead to a math book on the third draw. The bag composition changes after each draw based on whether a math or physics book is drawn, requiring careful tracking of conditional probabilities.
Define events $E_1, E_2, E_3, E_4$ for the first two books being (math, math), (math, phy), (phy, math), (phy, phy) respectively, and $A$ for the third book being math. Calculate $P(E_1) = \frac{3}{5} \times \frac{3}{7} = \frac{3}{7}$, $P(E_2) = \frac{3}{5} \times \frac{2}{7} = \frac{6}{35}$, $P(E_3) = \frac{2}{5} \times \frac{3}{7} = \frac{3}{10}$, $P(E_4) = \frac{2}{5} \times \frac{1}{7} = \frac{1}{10}$. Then find conditional probabilities $P(A|E_i)$ and apply Bayes' theorem: $P(E_4|A) = \frac{P(E_4)P(A|E_4)}{P(A)} = \frac{42}{347}$.
Correct Answer: 3