Probability 30271

The probability of an adverse drug reaction is 2.1%. How many people need to be prescribed for one patient to experience side effects with a chance of 90%?

Correct answer:

n = 0

Step-by-step explanation:

q = 2.1 % = \frac{2 . 1}{1 0 0} = 0.021 p_{2} = 90 % = \frac{9 0}{1 0 0} = \frac{9}{1 0} = 0.9 p_{2} = (1 n) \cdot q^{1} \cdot (1 - q)^{n - 1} p_{2} = n \cdot q \cdot (1 - q)^{n - 1} 0.9 = n \cdot 0.021 \cdot (1 - 0.021)^{n - 1} 0.9 = n \cdot 0.021 \cdot (1 - 0.021)^{n - 1} 42.857 = n \cdot 0.97 9^{n - 1} C_{1} (10) = (1 1 0) = \frac{1 0 !}{1 ! ( 1 0 - 1 ) !} = \frac{1 0}{1} = 10 p_{10} = (1 1 0) \cdot q^{1} \cdot (1 - q)^{(} 10 - 1) = 10 \cdot 0.02 1^{1} \cdot (1 - 0.021)^{(} 10 - 1) ≐ 0.1735 C_{1} (100) = (1 1 0 0) = \frac{1 0 0 !}{1 ! ( 1 0 0 - 1 ) !} = \frac{1 0 0}{1} = 100 p_{100} = (1 1 0 0) \cdot q^{1} \cdot (1 - q)^{(} 100 - 1) = 100 \cdot 0.02 1^{1} \cdot (1 - 0.021)^{(} 100 - 1) ≐ 0.2569 C_{1} (200) = (1 2 0 0) = \frac{2 0 0 !}{1 ! ( 2 0 0 - 1 ) !} = \frac{2 0 0}{1} = 200 p_{200} = (1 2 0 0) \cdot q^{1} \cdot (1 - q)^{(} 200 - 1) = 200 \cdot 0.02 1^{1} \cdot (1 - 0.021)^{(} 200 - 1) ≐ 0.0615 C_{1} (1000) = (1 1 0 0 0) = \frac{1 0 0 0 !}{1 ! ( 1 0 0 0 - 1 ) !} = \frac{1 0 0 0}{1} = 1000 p_{1000} = (1 1 0 0 0) \cdot q^{1} \cdot (1 - q)^{1000 - 1} = 1000 \cdot 0.02 1^{1} \cdot (1 - 0.021)^{1000 - 1} ≐ 1.3006 \cdot 1 0^{- 8} n = 0

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