A pair of fair dice is rolled. If sum of the numbers obtained is less than 5 then a letter is chosen at random from the letters of the word DIVIDE. If the sum is more than 9 then a letter is chosen at random from the letters of the word GARDEN. Otherwise a letter is chosen at random from the letters of the word INDIAN. If probability of getting a prime score when the selected letter is a vowel is ‘p’, then the integral part of 17p is

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answer is 7.

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Detailed Solution

Let $E_{1} : sum 〈5, E_{2} : 6 \leq sum \leq 9 and E_{3} : sum〉 9$
${PE}_{1} = \frac{1 + 2 + 3}{36} = \frac{1}{6}, {PE}_{3} = \frac{1}{6} \Rightarrow {PE}_{2} = \frac{24}{36} = \frac{4}{6}$
Let E be the event of choosing a vowel. Using Baye’s theorem
$P (\frac{E_{1}}{E}) = [\frac{\frac{1}{6} \times \frac{3}{6}}{\frac{1}{6} \times \frac{3}{6} + \frac{4}{6} \times \frac{3}{6} + \frac{1}{6} \times \frac{2}{6}}] = \frac{3}{17}$
Similarly $P (E_{2} / E) = 12 / 17 and P (E_{3} / E) = 2 / 17$
Required probability
$\begin{array}{l} P = \frac{3}{17} \times (\frac{1 + 2}{6}) + \frac{12}{17} \times (\frac{4 + 6}{24}) + \frac{2}{17} \times (\frac{2}{6}) \\ 17 P = [\frac{3}{2} + 5 + \frac{2}{3}] = \frac{43}{6} \Rightarrow 17 p = 7 \end{array}$