Assuming the dice rolls are independent, the probability of getting doubles is
![\dbinom61\dfrac16\cdot\dfrac16=\dfrac6{36}=\dfrac16](https://tex.z-dn.net/?f=%5Cdbinom61%5Cdfrac16%5Ccdot%5Cdfrac16%3D%5Cdfrac6%7B36%7D%3D%5Cdfrac16)
. (That is, there are 6 ways of getting doubles, and the probability of getting a given face is 1/6.)
So the event of getting doubles in a set number of
![n](https://tex.z-dn.net/?f=n)
trials is binomially distributed with
![p=\dfrac16](https://tex.z-dn.net/?f=p%3D%5Cdfrac16)
(and so
![q=1-p=\dfrac56](https://tex.z-dn.net/?f=q%3D1-p%3D%5Cdfrac56)
).
You're given that
![n=100](https://tex.z-dn.net/?f=n%3D100)
. If
![X](https://tex.z-dn.net/?f=X)
is a random variable representing the number of doubles obtained in
![n](https://tex.z-dn.net/?f=n)
trials, then the probability of getting
![X=x](https://tex.z-dn.net/?f=X%3Dx)
doubles is given by the PMF,
Answer:
361/900
Step-by-step explanation:
![\left(-\dfrac{1}{6}+0.6\left(-\dfrac{1}{3}\right)+1\right)^2=\left(-\dfrac{1}{6}-0.2+1\right)^2\\\\=\left(1-\left(\dfrac{1}{6}+\dfrac{1}{5}\right)\right)^2=\left(1-\dfrac{5+6}{6\cdot5}\right)^2=\left(\dfrac{19}{30}\right)^2=\boxed{\dfrac{361}{900}}](https://tex.z-dn.net/?f=%5Cleft%28-%5Cdfrac%7B1%7D%7B6%7D%2B0.6%5Cleft%28-%5Cdfrac%7B1%7D%7B3%7D%5Cright%29%2B1%5Cright%29%5E2%3D%5Cleft%28-%5Cdfrac%7B1%7D%7B6%7D-0.2%2B1%5Cright%29%5E2%5C%5C%5C%5C%3D%5Cleft%281-%5Cleft%28%5Cdfrac%7B1%7D%7B6%7D%2B%5Cdfrac%7B1%7D%7B5%7D%5Cright%29%5Cright%29%5E2%3D%5Cleft%281-%5Cdfrac%7B5%2B6%7D%7B6%5Ccdot5%7D%5Cright%29%5E2%3D%5Cleft%28%5Cdfrac%7B19%7D%7B30%7D%5Cright%29%5E2%3D%5Cboxed%7B%5Cdfrac%7B361%7D%7B900%7D%7D)
Answer:
I think it might be 14
Step-by-step explanation:
Answer:
12 boys
Step-by-step explanation:
Multiply the ratio numbers by 6 and you have
boys : girls = 2 : 3 = 12 : 18
There are 12 boys in the class.