Question

A gardener wanted to plant vegetables in his garden. Hence he bought 10 seeds of brinjal plant, 12 seeds of cabbage plant, and 8 seeds of radish plant. The shopkeeper assured him of germination probabilities of brinjal, cabbage, and radish to be 25%, 35%, and 40% respectively. But before he could plant the seeds, they got mixed up in the bag and he had to sow them randomly.

Calculate the probability of a randomly chosen seed to germinate.

Hint:

For calculating the total probability, use the law of total probability. Multiply the probability of each event by the probability of selecting each type of seed.

Answer 1

The dataset comprises 10 brinjal seeds, 12 cabbage seeds, and 8 radish seeds, totaling 30 seeds (\( 10 + 12 + 8 = 30 \)). The germination probabilities for each seed type are provided as follows: Brinjal: \( P(\text{Brinjal}) = 0.25 \), Cabbage: \( P(\text{Cabbage}) = 0.35 \), and Radish: \( P(\text{Radish}) = 0.40 \). The objective is to determine the overall probability of a randomly selected seed germinating. This calculation employs the law of total probability: \[ P(\text{Germinate}) = P(\text{Brinjal}) \cdot P(\text{Brinjal seed}) + P(\text{Cabbage}) \cdot P(\text{Cabbage seed}) + P(\text{Radish}) \cdot P(\text{Radish seed}) \] The probabilities of selecting each type of seed are: \( P(\text{Brinjal seed}) = \frac{10}{30} = \frac{1}{3} \), \( P(\text{Cabbage seed}) = \frac{12}{30} = \frac{2}{5} \), and \( P(\text{Radish seed}) = \frac{8}{30} = \frac{4}{15} \). Substituting these values into the formula yields: \[ P(\text{Germinate}) = 0.25 \cdot \frac{1}{3} + 0.35 \cdot \frac{2}{5} + 0.40 \cdot \frac{4}{15} \] The computation proceeds as: \[ P(\text{Germinate}) = \frac{0.25}{3} + \frac{0.70}{5} + \frac{1.60}{15} \] Summing these terms gives: \[ P(\text{Germinate}) = \frac{0.25}{3} + \frac{0.14}{1} + \frac{0.1067}{1} \] Consequently, the estimated probability of a seed germinating is: \[ P(\text{Germinate}) \approx 0.267 \]