Question

if a seed is planted, it has a 60% chance of growing into a healthy plant.
if 6 seeds are planted, what is the probability that exactly 3 dont grow?

Explanation:

Step1: Determine the probability of growing and not - growing

The probability of a seed growing \(p = 0.6\), so the probability of a seed not - growing \(q=1 - p=1 - 0.6 = 0.4\)

Step2: Use the binomial probability formula

The binomial probability formula is \(P(X = k)=C(n,k)\times p^{n - k}\times q^{k}\), where \(n\) is the number of trials, \(k\) is the number of "successes" (in this case, non - growing seeds), and \(C(n,k)=\frac{n!}{k!(n - k)!}\)
Here, \(n = 6\), \(k = 3\)
First, calculate the combination \(C(6,3)=\frac{6!}{3!(6 - 3)!}=\frac{6!}{3!3!}=\frac{6\times5\times4}{3\times2\times1}=20\)
Then, \(p^{n - k}=0.6^{6 - 3}=0.6^{3}=0.216\), \(q^{k}=0.4^{3}=0.064\)

Step3: Calculate the probability

\(P(X = 3)=C(6,3)\times0.6^{3}\times0.4^{3}\)
\(P(X = 3)=20\times0.216\times0.064\)
\(P(X = 3)=20\times0.013824\)
\(P(X = 3)=0.27648\)

Answer:

\(0.27648\)