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.

5 0

3 years ago

both brian and jermaine solve the problem. brian says the answer is $40,704. jermaine's answer is $4,604.

Studentka2010 [4]

Could you possibly be more specific?what math problem did they solve? i can help but just be a little more specific so i can answer your question

7 0

3 years ago

949 nearest 10 and 100

stellarik [79]

The nearest 10 is 50 and the the nearest hundred is 900

6 0

3 years ago

Read 2 more answers

What is the exterior angle of #9 ?

Alexxandr [17]

Answer:

Step-by-step explanation:

3x-12=2x

+12

3x=2x +12

-2x

x=12

4(12)+ 9

48+9

=55

6 0

3 years ago