Questions: If a seed is planted, it has a 70% chance of growing into a healthy plant. If 9 seeds are planted, what is the probability that exactly 4 don't grow?

If a seed is planted, it has a 70% chance of growing into a healthy plant. If 9 seeds are planted, what is the probability that exactly 4 don't grow?

Transcript text: If a seed is planted, it has a 70\% chance of growing into a healthy plant. If 9 seeds are planted, what is the probability that exactly 4 don't grow? $\square$

Solution

Solution Steps

Step 1: Define the Problem

We are tasked with finding the probability that exactly 4 out of 9 seeds do not grow into healthy plants, given that each seed has a $70\%$ chance of growing. This can be framed in terms of a binomial distribution where:

$n = 9$ (the number of trials, or seeds planted),
$p = 0.7$ (the probability of success, or a seed growing),
$q = 1 - p = 0.3$ (the probability of failure, or a seed not growing).

Step 2: Calculate the Required Probability

To find the probability that exactly 4 seeds do not grow, we need to determine the probability that exactly 5 seeds grow. This is given by the binomial probability formula:

\[ P(X = x) = \binom{n}{x} \cdot p^x \cdot q^{n-x} \]

In our case, we need to calculate $P(X = 5)$ (where $x = 5$ represents the number of seeds that grow):

\[ P(X = 5) = \binom{9}{5} \cdot (0.7)^5 \cdot (0.3)^{4} \]

Step 3: Compute the Probability

Using the binomial coefficient $\binom{9}{5} = \frac{9!}{5!(9-5)!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$, we can substitute into the formula:

\[ P(X = 5) = 126 \cdot (0.7)^5 \cdot (0.3)^{4} \]

Calculating the powers:

\[ (0.7)^5 \approx 0.16807 \quad \text{and} \quad (0.3)^{4} = 0.0081 \]

Now substituting these values back into the equation:

\[ P(X = 5) = 126 \cdot 0.16807 \cdot 0.0081 \approx 0.1715 \]