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vfiekz [6]
3 years ago
6

A scientist inoculates​ mice, one at a​ time, with a disease germ until he finds 3 that have contracted the disease. If the prob

ability of contracting the disease is two sevenths ​, what is the probability that 8 mice are​ required? The probability that that 8 mice are required is nothing.
Mathematics
1 answer:
UkoKoshka [18]3 years ago
5 0

Answer:

The probability that 8 mice are​ required is 0.2428.

Step-by-step explanation:

Given : A scientist inoculates​ mice, one at a​ time, with a disease germ until he finds 3 that have contracted the disease. If the probability of contracting the disease is two sevenths.

To find : What is the probability that 8 mice are​ required? The probability that that 8 mice are required is nothing ?

Solution :

Applying binomial distribution,

P(X=r)=^nC_r p^rq^{n-r}

Where, p is the probability of success p=\frac{2}{7}

q is the probability of failure q=1-p, q=1-\frac{2}{7}=\frac{5}{7}

n is total number of trials n=8

r=3

Substitute the values,

P(X=3)=^8C_3 (\frac{2}{7})^3 (\frac{5}{7})^{8-3}

P(X=3)=\frac{8!}{3!5!}\times \frac{8}{343}\times (\frac{5}{7})^{5}

P(X=3)=\frac{8\times 7\times 6}{3\times 2\times 1}\times \frac{8}{343}\times \frac{3125}{16807}

P(X=3)=0.2428

Therefore, the probability that 8 mice are​ required is 0.2428.

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