In a certain biological experiment some animals are analysed. Each

time the experiment is repeated a different animal is analysed, so

that no more than one animal is used is used in each repetition.

Knowing that the probability that the experiment is successful is of

40%40\%, determine the number of animals necessary so that the

probability of obtaining at least one successful experiment isn’t

inferior to .95.95.

I did:

(the numbers in red are the number of experiments)

Probabilities of getting at least one successful experiment:

1 experiment: 1−(.6)1-(.6) \\

2 experiments: 1−(.6.6)1-(.6.6) \\

3 experiments: 1−(.6.6.6)1-(.6.6.6)

And so on… so I made a formula: 1−(.6)k=p1 – (.6)^k = p

(k is the number os experiments and p is the probability of being successful at least once)

So the answer to this problem would be:

1−(.6)k=.951 – (.6)^k = .95

I put this on my calculator and the intersection is 5.8…≈65.8… \approx 6, which my book says is the correct result.

My questions:

How do I solve 1−(.6)k=.951 – (.6)^k = .95 step by step, instead of using the calculator?

Is my solution correct?

Is there a better/faster way of solving this problem? If yes, how?

Thanks

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@carmichael561 Somewhat…

– SilenceOnTheWire

2 days ago

@carmichael561 Not much.

– SilenceOnTheWire

2 days ago

@carmichael561 Are you suggesting that the solution has something to do with the normalcdf?

– SilenceOnTheWire

2 days ago

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1 Answer

1

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If the probability of success is .4, the probability of failure is .6. The probability of no success in n trials is .6n.6^n. The probability of “at least one success” in n trials is 1−.6n1- .6^n. So you want n such that 1−.6n=.951- .6^n= .95 as you say. I’m not sure what you mean by “without a calculator”. You need to use “logarithms” to solve that since logarithms are the inverse function to exponentials. .6n=1−.95=0.05.6^n= 1- .95= 0.05 so that log(.6n)=nlog(.6)=log(0.05)log(.6^n)= n log(.6)= log(0.05) and n=log(0.05)log(.6)”n= \frac{log(0.05)}{log(.6)”}. The right side of that is not an integer so round up to an integer. Those of us who learned math in the years “B.C.” (before calculators) used tables of logarithms. It is possible to “calculate” logarithms with paper and pencil but I would not recommend it.

Right, I understand your answer. Also, when I say “without a calculator” I mean that the use of calculators in my course is somewhat strict… we can use them but most of the calculations have to be done by hand (ridiculous, but what can I do?) so since there if a fine line as to when calculators can be used or not, I prefer to solve everything without a calculator whenever possible. That’s why.

– SilenceOnTheWire

2 days ago

1

You can write the answer as n \ge \frac{\log .05}{\log .6}n \ge \frac{\log .05}{\log .6} and that should be considered final enough. If you have to estimate without a calculator you can go n = \frac{\log_{10} .05}{\log_{10} .6} = \frac{\log_{10} 5 – 2}{\log_{10} 6 – 1}n = \frac{\log_{10} .05}{\log_{10} .6} = \frac{\log_{10} 5 – 2}{\log_{10} 6 – 1}. and … lessee 5^3 \approx 100= 10^25^3 \approx 100= 10^2 so 5 \approx 10^{2/3}5 \approx 10^{2/3} and 6^2 = 36; 6^4 \approx 1000=10^36^2 = 36; 6^4 \approx 1000=10^3 so 6 \approx 10^{3/4}6 \approx 10^{3/4} so n \approx \frac{\frac 23 – 2}{\frac 34 – 1}\approx 5.3n \approx \frac{\frac 23 – 2}{\frac 34 – 1}\approx 5.3 so about 6 animals… But that’s really rough and really pointless in this day of calculators. n \ge \frac{\log .05}{\log .6}n \ge \frac{\log .05}{\log .6} really ought to be enough.

– fleablood

2 days ago

Find out if expressing as a rational involving logarithms is good enough. It really ought to be. Then ask if it’s okay to use a calculator for roots and logarithms. That really ought to be too.

– fleablood

2 days ago