Suppose you purchase a bundle of 10 bare-root rhubarb plants. The sales clerk tells you that 5% of these plants will die before producing any rhubarb. Assume that the bundle is a random sample of plants and that the sales clerk’s statement is accurate. Let Y = the number of plants that die before producing any rhubarb. Use the binomial probability formula to find P(Y = 1). Interpret this result in context.

For each plant, there are only two possible outcomes. Either they are going to die before producing any rhubarb, or they are not. This means that we can solve this problem using the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinatios of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

In this problem we have that:

There are 10 plants, so [tex]n = 10[/tex].

5% of these plants will die before producing any rhubarb. This means that [tex]p = 0.05[/tex].

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(Y = 1) = C_{10,1}.(0.05)^{1}.(0.95)^{9} = 0.3151[/tex]

[tex]P(Y = 1) = 0.3151[/tex]. This means that there is a 31.51% probability that exactly one plant in the sample dies before producing any rhubarb.