Question

One thousand independent rolls of a fair die will be made. Compute an approximation to the probability that the number 6 will appear between 150 and 200 times inclusively. If the number 6 appears exactly 200 times, find the probability that the number 5 will appear less than 150 times?

Answer 1

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)=$\frac{1}{6}$

Let X be the random variable  for the number of sixes  in the 1000 rolls of die.

Then, $X\sim Binom(1000,\frac{1}{6})$

Since, n is very large,the binomial random variable can be approximated as normal random variable.

Mean,$\mu=np=1000\times \frac{1}{6}=166.67$

Variance=$\sigma^2=np(1-p)=1000\times \farc{1}{6}\times (1-\frac{1}{6})=1000\times \frac{5}{36}=138.89$

$X\sim N(166.67,138.89)$

$P(150\leq X\leq 200)=P[\frac{150-166.67}{11.79}\leq \frac{X-\mu}{\sigma}\leq \frac{200-166.67}{11.79}]$

=$P[-1.41\leq Z\leq 2.83]=P[Z\leq 2.83]-P[Z$

=$\phi(2.83)-\phi(-1.41)$

=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)=$\frac{1}{5}$

Let Y be the random variable denoting the number of times  5 coming up in 800 rolls.

Then, $Y\sim bin(800,\frac{1}{5})$

Mean,$\mu=np=800\times \frac{1}{5}=160$

Variance, $\sigma^2=np(1-p)=800\times \frac{1}{5}(1-\frac{1}{5})=128$

$Y\sim N(160,128)$ because n is large

$P(Y$

$P(Y$

Hence, the probability that the number 5 will appear less than 150 times given that 6 appeared exactly 200 times=0.19