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?

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)=

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

Then,

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

Mean,

Variance=

=

=

=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)=

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

Then,

Mean,

Variance,

because n is large

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

it would not be increasing by nothing that i can see probably by 25 percent that is all i can sat\y for now if i have any thing else to say i will edit this.

step-by-step explanation: