Let A be set of three odd different digits and B be the set of three even non‑zero different digits. A five digit number is made at random out of these digits, all digits being different and p is the probability that no odd digit falls between any two even digits in the number. Also out of all the mappings defined from the set A → B, a mapping is randomly selected and the probability that selected mapping is bijective is q. If \frac{p}{q} = \frac{7 k}{40}, where k \in R then find k.

Engineering

Mathematics

Conditional Probability

Question

Let A be set of three odd different digits and B be the set of three even non‑zero different digits. A five digit number is made at random out of these digits, all digits being different and p is the probability that no odd digit falls between any two even digits in the number. Also out of all the mappings defined from the set A → B, a mapping is randomly selected and the probability that selected mapping is bijective is q. If $\frac{p}{q} = \frac{7 k}{40}$ , where k $\in$ R then find k.

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Five digit number – (favourable curve)

$= \underset{2 odd, 3 even}{\underset{⏟}{^{3} C_{2} \times 4! \times 2!}} + \underset{3 odd, 2 even}{\underset{⏟}{^{3} C_{2} \times 3! \times 3!}}$

= 3 × 24 × 2 + 3 × 6 × 6 = 144 + 108 = 252

$∴$ Probability $p = \frac{252}{^{6} C_{5} \times 5!} = \frac{252}{720} = \frac{7}{20}$

and Probability, $q = \frac{3!}{3^{3}} = \frac{6}{27} = \frac{2}{9}$

$∴ \frac{p}{q} = \frac{7 \times 9}{40} \Rightarrow k = 9 .$