A seed company claims that 90% of its bean seeds will germinate. If seven of these seeds
are planted in warm, moist, soil, what is the probability that exactly 6 of them will germinate?
Round answer to nearest ten-thousandth (4 places after decimal)
prob(good) = .9
prob(bad) = .1
you want prob(6 of 7 to be good)
= C(7,6)(.9)^6 (.1)
= ..
you push the buttons
0.3720
To find the probability of exactly 6 of the seeds germinating, we will use the binomial probability formula.
The binomial probability formula is given by:
P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)
Where:
- P(X = k) is the probability of exactly k successes
- C(n, k) is the number of combinations (n choose k)
- p is the probability of success for each trial
- n is the total number of trials
In this case, we have:
- n = 7 (total number of seeds)
- k = 6 (the desired number of germinated seeds)
- p = 0.9 (probability of each seed germinating)
First, let's calculate C(n, k) using the combination formula:
C(n, k) = n! / (k! * (n - k)!)
C(7, 6) = 7! / (6! * (7 - 6)!)
= 7! / (6! * 1!)
= 7
Now, let's substitute the values into the binomial probability formula:
P(X = 6) = C(7, 6) * (0.9)^6 * (1 - 0.9)^(7 - 6)
= 7 * 0.9^6 * 0.1^1
≈ 7 * 0.5314 * 0.1
≈ 0.37198
Rounding to four decimal places, the probability that exactly 6 of the seeds will germinate is approximately 0.3720.