Mimi plans on growing tomatoes in her garden. She has 20 cherry tomato seeds. Based on her experience, the probability of a seed turning into a seedling is 0.60.
(a) Let X be the number of seedlings that Mimi gets. As we know, the distribution of X is a binomial probability distribution. What is the number of trials (n), probability of successes (p) and probability of
failures (q), respectively?
(b) Find the probability that she gets at least 15 cherry tomato seedlings. (round the answer to 3 decimal places)
Solution
a. There are 20 cherry tomato seeds.
Let $X$ denote the number of seedlings that Mimi gets.
The the probability of a seed turning into a seedling is 0.60, i.e. $p = r p$.
The number of trials $n = 20$.
The probability of success is $p = 0.6$.
The probability of failure is $q = 1- p = 1 - 0.6 = 0.4$.
The random variable $X$ follows Binomial distribution with parameters $n = 20$ and $p=0.6$.
Thus $X\sim B(20, 0.6)$.
The probability mass function of $X$ is
$$
P(X=x) = \binom{20}{x} (0.6)^x (1-0.6)^{20-x}, \; x=0,1,\cdots, 20.
$$
b. The probability that she gets at least 15 cherry tomato seedlings is
$$ \begin{aligned} P(X\geq 15) &= P(X=15)+P(X=16)+P(X=17)+P(X=18)+P(X=19)+P(X=20)\\ &=\binom{20}{15} (0.6)^{15} (1-0.6)^{20-15}+\binom{20}{16} (0.6)^{16} (1-0.6)^{20-16}\\ & \quad +\binom{20}{17} (0.6)^{17} (1-0.6)^{20-17}+\binom{20}{18} (0.6)^{18} (1-0.6)^{20-18}\\ &\quad +\binom{20}{19} (0.6)^{19} (1-0.6)^{20-19}+\binom{20}{20} (0.6)^{20} (1-0.6)^{20-20}\\ &=0.075+0.035+0.012+0.003+0+0\\ &=0.125 \end{aligned} $$
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
