A seed packet contains 4 white, 5 black, and 6 red seeds. If 3 seeds are drawn at random simultaneously without replacement, the probability that all three drawn seeds are of pairwise different colours is expressed in simplest fractional form as $\frac{p}{q}$. The value of $(p + q)$ is ________ (answer in integer).
Answer Key:
115
Step-by-Step Solution & Explanation
1. Total seeds in packet $= 4 + 5 + 6 = 15$.
2. Total possible ways to choose 3 seeds:
$n(S) = \binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455$
3. Ways to draw exactly 1 white, 1 black, and 1 red seed:
$n(E) = \binom{4}{1} \times \binom{5}{1} \times \binom{6}{1} = 4 \times 5 \times 6 = 120$
4. Probability:
$P = \frac{120}{455} = \frac{24}{91}$
Since $\gcd(24, 91) = 1$, $p = 24$ and $q = 91$.
5. Value of $p + q = 24 + 91 = 115$.