Question

a drug is effective in nine cases out of ten. if it is tested on 12 patients, what is the probability that it will be ineffective in more than two cases?

Answer 1

Probability that the drug will be ineffective in more than two cases is 0,68.

Probability

Drug is effective in 9 cases out of 10, so the drug:

• Effective at 9/10
• Ineffective at 1/10

If the drug is tested on 12 patients, then there are 13 possibilities, namely:

• Drug effective in 0 patients, ineffective in 12 patients
• Drug effective in 1 patient, ineffective in 11 patients
• Drug effective in 2 patients, ineffective in 10 patients
• .
• .
• .
• Drug effective in 12 patients, ineffective in 0 patients

In the problem, asked to find the probability of the drug being ineffective in more than 2 patients, there are 10 possibilities, namely

• Drug effective in 9 patients, ineffective in 3 patients
• Drug effective in 8 patients, ineffective in 4 patients
• .
• .
• .
• Drug effective in 0 patients, ineffective in 12 patients

The way to find out the possibility of a drug being effective in 0-9 patients is:

P(effective ≤ 9) = P(effective 9) + P(effective 8) + ... + P(effective 0)

However, quite a lot is calculated if you use the formula above, less will be calculated if you use the method below:

P(effective ≤ 9) = 1 - ( P(effective 10) + P(effective 11) + P(effective 12) )

Let's try to start by calculating P(effective 10). P(effective 10) means the probability of being effective for 10 people and ineffective for 2 people. Then the effective probability is raised to the power of 10 and multiplied by the ineffective probability raised to the power of 2.

P(effective 10) = $(\frac{9}{10})^{10} .(\frac{1}{10} )^2$

P(effective 10) = $\frac{9^{10} }{10^{12} }$

P(effective 10) = $\frac{3486784401}{1000000000000}$

So does P(effective 11) and P(effective 12)

P(effective 11) = $(\frac{9}{10} )^{11}.(\frac{1}{10})^{1}$

P(effective 11) = $\frac{9^{11} }{10^{12} }$

P(effective 11) = $\frac{3486784401}{1000000000000}$

P(effective 12) = $(\frac{9}{10}) ^{12} .( \frac{1}{10} )^0$

P(effective 12) = $\frac{9^{12}}{10^{12}}$

P(effective 12) = $\frac{282429536481}{1000000000000}$

So:

P(effective ≤ 9) = 1 - ( P(effective 10) + P(effective 11) + P(effective 12) )

P(effective ≤ 9) = 1 - ( $\frac{3486784401}{1000000000000}$+ $\frac{3486784401}{1000000000000}$+ $\frac{282429536481}{1000000000000}$)

P(effective ≤ 9) = 1 - $\frac{317297380491}{1000000000000}$

P(effective ≤ 9) = 1 - 0,317297380491

P(effective ≤ 9) = 0,682702619509 ≈ 0,68

Learn more about probability here: https://brainly.com/question/16484393

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