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)=[tex]\frac{1}{6}[/tex]
Let X be the random variable for the number of sixes in the 1000 rolls of die.
Then, [tex]X\sim Binom(1000,\frac{1}{6})[/tex]
Since, n is very large,the binomial random variable can be approximated as normal random variable.
Mean,[tex]\mu=np=1000\times \frac{1}{6}=166.67[/tex]
Variance=[tex]\sigma^2=np(1-p)=1000\times \farc{1}{6}\times (1-\frac{1}{6})=1000\times \frac{5}{36}=138.89[/tex]
[tex]X\sim N(166.67,138.89)[/tex]
[tex]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}][/tex]
=[tex]P[-1.41\leq Z\leq 2.83]=P[Z\leq 2.83]-P[Z<-1.41][/tex]
=[tex]\phi(2.83)-\phi(-1.41)[/tex]
=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)=[tex]\frac{1}{5}[/tex]
Let Y be the random variable denoting the number of times 5 coming up in 800 rolls.
Then, [tex]Y\sim bin(800,\frac{1}{5})[/tex]
Mean,[tex]\mu=np=800\times \frac{1}{5}=160[/tex]
Variance, [tex]\sigma^2=np(1-p)=800\times \frac{1}{5}(1-\frac{1}{5})=128[/tex]
[tex]Y\sim N(160,128)[/tex] because n is large
[tex]P(Y<150)=P(\frac{Y-\mu}{\sigma}<\frac{150-160}{11.31})[/tex]
[tex]P(Y<150)=P(Z<-0.884)=\phi(-0.88)=0.18943[/tex]
Hence, the probability that the number 5 will appear less than 150 times given that 6 appeared exactly 200 times=0.19
