Respuesta :
Given:
p = 90% = 0.9, the probability that the seed grows into a healthy plant.
Therefore
q = 1 - p = 0.1, the probability that the seed will not grow into a healthy plant.
n = 7, the number of seeds planted.
Let r = 4, expected number of seeds that will grow.
Use the Binomial distribution to determine
P(4 of 7 seeds will grow) = ₇C₄ p⁴q⁽⁷⁻⁴)
= 35*(0.9⁴)*(0.1³)
= 0.023
P(4 of 7 seeds will not grow) = 1 - P(4 of 7 seeds will grow)
= 1 - 0.023
= 0.977 or 97.7%
Answer: 97.7%
p = 90% = 0.9, the probability that the seed grows into a healthy plant.
Therefore
q = 1 - p = 0.1, the probability that the seed will not grow into a healthy plant.
n = 7, the number of seeds planted.
Let r = 4, expected number of seeds that will grow.
Use the Binomial distribution to determine
P(4 of 7 seeds will grow) = ₇C₄ p⁴q⁽⁷⁻⁴)
= 35*(0.9⁴)*(0.1³)
= 0.023
P(4 of 7 seeds will not grow) = 1 - P(4 of 7 seeds will grow)
= 1 - 0.023
= 0.977 or 97.7%
Answer: 97.7%
Answer:
0.098415
Step-by-step explanation:
The way the question is worded is odd and you should probably actually include “don’t grow into a healthy plant” or something. You can use the binomial distribution. Suppose ∼Bin(6,0.1) so the probability that k trees don’t grow into a healthy plant is
Pr( = ) = (6 / ) 0.16^ 0.9^6−
Then the probability that two don’t grow into healthy plants is
Pr( = 2) = (6/2) 0.1^2 0.9^4
this is equal to 0.098415
