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

4 years ago

Find the slope of the line through each pair of points (-12,-11),(13,-6)

Mariana [72]

5/25 or 1/5

Can u mark me brainliest

7 0

3 years ago

Y is inversely proportional to the square of x. If Y = 9 when x = 2, then what is y, when x is 10?

a_sh-v [17]

Answer:

y = 0.36

Step-by-step explanation:

Given y is inversely proportional to x² then the equation relating them is

y = $\frac{k}{x^2}$ ← k is the constant of variation

To find k use the condition y = 9 when x = 2 , then

9 = $\frac{k}{2^2}$ = $\frac{k}{4}$ ( multiply both sides by 4 )

36 = k

y = $\frac{36}{x^2}$ ← equation of variation

When x = 10 , then

y = $\frac{36}{10^2}$ = $\frac{36}{100}$ = 0.36

5 0

3 years ago