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Question Fertilizer boosts seed germination by 75%. Find probability that at most 4 out of 7 randomly selected seeds germinate.

Studdy Solution

STEP 1

Assumptions
1. The probability of a single seed germinating with the aid of the fertilizer is 75%75\% or 0.750.75.
2. The total number of seeds planted and randomly selected is 77.
3. "At most 4 germinate" means we are looking for the probability of 0,1,2,3,0, 1, 2, 3, or 44 seeds germinating.
4. The events are independent, and the number of successes in a fixed number of trials follows a binomial distribution.

STEP 2

The probability of exactly kk successes in nn trials in a binomial distribution is given by the formula:
P(X=k)=(nk)pk(1−p)n−kP(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
where P(X=k)P(X = k) is the probability of kk successes, (nk)\binom{n}{k} is the binomial coefficient, pp is the probability of success on a single trial, and (1−p)(1-p) is the probability of failure on a single trial.

STEP 3

First, calculate the probability of 00 seeds germinating using the binomial probability formula.
P(X=0)=(70)(0.75)0(1−0.75)7−0P(X = 0) = \binom{7}{0} (0.75)^0 (1-0.75)^{7-0}

STEP 4

Calculate the binomial coefficient and the probabilities.
P(X=0)=1⋅1⋅(0.25)7P(X = 0) = 1 \cdot 1 \cdot (0.25)^7

STEP 5

Compute the probability of 00 seeds germinating.
P(X=0)=(0.25)7P(X = 0) = (0.25)^7

STEP 6

Calculate the probability of 11 seed germinating.
P(X=1)=(71)(0.75)1(1−0.75)7−1P(X = 1) = \binom{7}{1} (0.75)^1 (1-0.75)^{7-1}

STEP 7

Compute the binomial coefficient and the probabilities.
P(X=1)=7⋅0.75⋅(0.25)6P(X = 1) = 7 \cdot 0.75 \cdot (0.25)^6

STEP 8

Calculate the probability of 22 seeds germinating.
P(X=2)=(72)(0.75)2(1−0.75)7−2P(X = 2) = \binom{7}{2} (0.75)^2 (1-0.75)^{7-2}

STEP 9

Compute the binomial coefficient and the probabilities.
P(X=2)=21⋅(0.75)2⋅(0.25)5P(X = 2) = 21 \cdot (0.75)^2 \cdot (0.25)^5

STEP 10

Calculate the probability of 33 seeds germinating.
P(X=3)=(73)(0.75)3(1−0.75)7−3P(X = 3) = \binom{7}{3} (0.75)^3 (1-0.75)^{7-3}

STEP 11

Compute the binomial coefficient and the probabilities.
P(X=3)=35⋅(0.75)3⋅(0.25)4P(X = 3) = 35 \cdot (0.75)^3 \cdot (0.25)^4

STEP 12

Calculate the probability of 44 seeds germinating.
P(X=4)=(74)(0.75)4(1−0.75)7−4P(X = 4) = \binom{7}{4} (0.75)^4 (1-0.75)^{7-4}

STEP 13

Compute the binomial coefficient and the probabilities.
P(X=4)=35⋅(0.75)4⋅(0.25)3P(X = 4) = 35 \cdot (0.75)^4 \cdot (0.25)^3

STEP 14

Now, add the probabilities of 0,1,2,3,0, 1, 2, 3, and 44 seeds germinating to find the total probability of at most 44 seeds germinating.
P(X≤4)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

STEP 15

Substitute the calculated probabilities into the equation.
P(X≤4)=(0.25)7+7⋅0.75⋅(0.25)6+21⋅(0.75)2⋅(0.25)5+35⋅(0.75)3⋅(0.25)4+35⋅(0.75)4⋅(0.25)3P(X \leq 4) = (0.25)^7 + 7 \cdot 0.75 \cdot (0.25)^6 + 21 \cdot (0.75)^2 \cdot (0.25)^5 + 35 \cdot (0.75)^3 \cdot (0.25)^4 + 35 \cdot (0.75)^4 \cdot (0.25)^3

STEP 16

Calculate each term to at least four decimal places.
P(X≤4)≈0.0002+0.0073+0.0537+0.1780+0.2966P(X \leq 4) \approx 0.0002 + 0.0073 + 0.0537 + 0.1780 + 0.2966

STEP 17

Add the probabilities to find the total probability.
P(X≤4)≈0.0002+0.0073+0.0537+0.1780+0.2966P(X \leq 4) \approx 0.0002 + 0.0073 + 0.0537 + 0.1780 + 0.2966

STEP 18

Compute the sum.
P(X≤4)≈0.5358P(X \leq 4) \approx 0.5358

STEP 19

Round the answer to two decimal places as required by the problem.
P(X≤4)≈0.54P(X \leq 4) \approx 0.54
The probability that at most 4 of the 7 seeds germinate is approximately 0.540.54.

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