Seeds

The germination of seeds of a certain species of carrot is 96%. What is the probability that at least 25 seeds out of 30 will germinate?

Final Answer:

p = 0.9989

C_{25} (30) = (2 5 3 0) = \frac{3 0 !}{2 5 ! ( 3 0 - 2 5 ) !} = \frac{3 0 \cdot 2 9 \cdot 2 8 \cdot 2 7 \cdot 2 6}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 142506 C_{26} (30) = (2 6 3 0) = \frac{3 0 !}{2 6 ! ( 3 0 - 2 6 ) !} = \frac{3 0 \cdot 2 9 \cdot 2 8 \cdot 2 7}{4 \cdot 3 \cdot 2 \cdot 1} = 27405 C_{27} (30) = (2 7 3 0) = \frac{3 0 !}{2 7 ! ( 3 0 - 2 7 ) !} = \frac{3 0 \cdot 2 9 \cdot 2 8}{3 \cdot 2 \cdot 1} = 4060 C_{28} (30) = (2 8 3 0) = \frac{3 0 !}{2 8 ! ( 3 0 - 2 8 ) !} = \frac{3 0 \cdot 2 9}{2 \cdot 1} = 435 C_{29} (30) = (2 9 3 0) = \frac{3 0 !}{2 9 ! ( 3 0 - 2 9 ) !} = \frac{3 0}{1} = 30 C_{30} (30) = (3 0 3 0) = \frac{3 0 !}{3 0 ! ( 3 0 - 3 0 ) !} = \frac{1}{1} = 1 k = 96 % = \frac{9 6}{1 0 0} = \frac{2 4}{2 5} = 0.96 q = 1 - k = 1 - 0.96 = \frac{1}{2 5} = 0.04 n = 30 p_{25} = (2 5 n) \cdot k^{25} \cdot q^{5} = 142506 \cdot 0.9 6^{25} \cdot 0.0 4^{5} ≐ 0.0053 p_{26} = (2 6 n) \cdot k^{26} \cdot q^{4} = 27405 \cdot 0.9 6^{26} \cdot 0.0 4^{4} ≐ 0.0243 p_{27} = (2 7 n) \cdot k^{27} \cdot q^{3} = 4060 \cdot 0.9 6^{27} \cdot 0.0 4^{3} ≐ 0.0863 p_{28} = (2 8 n) \cdot k^{28} \cdot q^{2} = 435 \cdot 0.9 6^{28} \cdot 0.0 4^{2} ≐ 0.2219 p_{29} = (2 9 n) \cdot k^{29} \cdot q^{1} = 30 \cdot 0.9 6^{29} \cdot 0.0 4^{1} ≐ 0.3673 p_{30} = (3 0 n) \cdot k^{30} \cdot q^{0} = 1 \cdot 0.9 6^{30} \cdot 0.0 4^{0} ≐ 0.2939 p = p_{25} + p_{26} + p_{27} + p_{28} + p_{29} + p_{30} = 0.0053 + 0.0243 + 0.0863 + 0.2219 + 0.3673 + 0.2939 = 0.9989

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