Jerry's Seed Palace claims that 80% of its lima bean seeds will germinate. Suppose the company's claim is true. Callie buys a packet with 20 lima bean seeds from Jerry's Seed Palace and plants them in her garden. What is the probability that exactly 14 seeds will germinate?

A. 0.1091
B. 0.8909
C. 0.1958
D. 0.8042
E. 0.9133

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Answer:

A

Step-by-step explanation:

This is a Bernoulli trial problem.

Since the probability that the seed will germinate is 80% or 0.8, the probability that they won’t germinate is 0.2

Now let’s say the probability to germinate is p and the probability not to is a, we can now set up the Bernoulli equation in this regard. Since we are proposing that exactly 14 will germinate, it means 6 are not expected to germinate

P(14) = 20C14 (0.8)^14 (0.2)^6 = 0.109099700973

Option A is the right answer

You can use binomial distribution here.

The probability that exactly 14 seeds will germinate is given by:

Option A: 0.1091 is correct

Given that:

  • Company claims that 80% of lima bean seeds will germinate.
  • Callie buys 20 lima beans

To find:

Probability that exactly 14 seeds will germinate.

Using binomial distribution to get the needed probability:

We use binomial distribution when we have to track success and failure of multiple objects(multiple Bernoulli trials)

Let X be the random variable tracking the number of seeds that will germinate out of 20 seeds. The success probability is 80% or 0.8(showing probability of germination)

Thus. [tex]X \sim B(20, 0.8)[/tex]

The probability of X = x is given by:

[tex]P(X = x) = \: ^{20}C_x (0.8)^x (1-0.8)^{20-x}[/tex]

Thus:

[tex]P(X = 14) = \: ^{20}C_{14} (0.8)^{14} (1-0.8)^{20-14} = 38760 \times 0.04398 \times 0.000064 = 0.109\\P(X = 14) = 0.10909.. \approx 0.1091[/tex]

The probability that exactly 14 seeds will germinate is given by:

Option A: 0.1091 is correct

Learn more about binomial distribution here:

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