Answer:
0.19
Step-by-step explanation:
We are given that
Number of rolls of die=n=1000
Let the event of six coming up be success.Then, in each trial , the probability of success =p=P(success)=P(6)=![\frac{1}{6}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B6%7D)
Let X be the random variable for the number of sixes in the 1000 rolls of die.
Then, ![X\sim Binom(1000,\frac{1}{6})](https://tex.z-dn.net/?f=X%5Csim%20Binom%281000%2C%5Cfrac%7B1%7D%7B6%7D%29)
Since, n is very large,the binomial random variable can be approximated as normal random variable.
Mean,![\mu=np=1000\times \frac{1}{6}=166.67](https://tex.z-dn.net/?f=%5Cmu%3Dnp%3D1000%5Ctimes%20%5Cfrac%7B1%7D%7B6%7D%3D166.67)
Variance=![\sigma^2=np(1-p)=1000\times \farc{1}{6}\times (1-\frac{1}{6})=1000\times \frac{5}{36}=138.89](https://tex.z-dn.net/?f=%5Csigma%5E2%3Dnp%281-p%29%3D1000%5Ctimes%20%5Cfarc%7B1%7D%7B6%7D%5Ctimes%20%281-%5Cfrac%7B1%7D%7B6%7D%29%3D1000%5Ctimes%20%5Cfrac%7B5%7D%7B36%7D%3D138.89)
![X\sim N(166.67,138.89)](https://tex.z-dn.net/?f=X%5Csim%20N%28166.67%2C138.89%29)
![P(150\leq X\leq 200)=P[\frac{150-166.67}{11.79}\leq \frac{X-\mu}{\sigma}\leq \frac{200-166.67}{11.79}]](https://tex.z-dn.net/?f=P%28150%5Cleq%20X%5Cleq%20200%29%3DP%5B%5Cfrac%7B150-166.67%7D%7B11.79%7D%5Cleq%20%5Cfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%5Cleq%20%5Cfrac%7B200-166.67%7D%7B11.79%7D%5D)
=![P[-1.41\leq Z\leq 2.83]=P[Z\leq 2.83]-P[Z](https://tex.z-dn.net/?f=P%5B-1.41%5Cleq%20Z%5Cleq%202.83%5D%3DP%5BZ%5Cleq%202.83%5D-P%5BZ%3C-1.41%5D)
=![\phi(2.83)-\phi(-1.41)](https://tex.z-dn.net/?f=%5Cphi%282.83%29-%5Cphi%28-1.41%29)
=0.9977-0.0793=0.9184
Thus, the probability that the number 6 appears between 150 to 200 times=0.92
Now, given that 6 appears exactly 200 times .
Therefore, other number appear in other 800 rolls .
We have to find the probability that the number 5 will appear less than 150 times.
Therefore, for 800 rolls, let the event of 5 coming up be success.
Then , p=P(success)=P(5)=![\frac{1}{5}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B5%7D)
Let Y be the random variable denoting the number of times 5 coming up in 800 rolls.
Then, ![Y\sim bin(800,\frac{1}{5})](https://tex.z-dn.net/?f=Y%5Csim%20bin%28800%2C%5Cfrac%7B1%7D%7B5%7D%29)
Mean,![\mu=np=800\times \frac{1}{5}=160](https://tex.z-dn.net/?f=%5Cmu%3Dnp%3D800%5Ctimes%20%5Cfrac%7B1%7D%7B5%7D%3D160)
Variance, ![\sigma^2=np(1-p)=800\times \frac{1}{5}(1-\frac{1}{5})=128](https://tex.z-dn.net/?f=%5Csigma%5E2%3Dnp%281-p%29%3D800%5Ctimes%20%5Cfrac%7B1%7D%7B5%7D%281-%5Cfrac%7B1%7D%7B5%7D%29%3D128)
because n is large
![P(Y](https://tex.z-dn.net/?f=P%28Y%3C150%29%3DP%28%5Cfrac%7BY-%5Cmu%7D%7B%5Csigma%7D%3C%5Cfrac%7B150-160%7D%7B11.31%7D%29)
![P(Y](https://tex.z-dn.net/?f=P%28Y%3C150%29%3DP%28Z%3C-0.884%29%3D%5Cphi%28-0.88%29%3D0.18943)
Hence, the probability that the number 5 will appear less than 150 times given that 6 appeared exactly 200 times=0.19