Question

A fair die is rolled 1000 times. Let A be the event that the number of 6’s is in the interval[150,200], and B the event that the number of 5’s is exactly 200. (a) Approximate P(A).(b) Approximate P(A|B).

Answer 1

n=

1000

p=probability of a number=

1/6

here mean of distribution=μ=np=		166.67
and standard deviation σ=sqrt(np(1-p))=		11.79
for normal distribution z score =(X-μ)/σx

therefore from normal approximation of binomial distribution and continuity correction:

a)

P(A) =P(149.5<X<200.5)=P((149.5-166.67)/11.785)<Z<(200.5-166.67)/11.785)=P(-1.46<Z<2.87)=0.9979-0.0721=0.9258

b)

n=

800

p(A|B)=probability of 5 given 6 is not in them=1/5=

0.20

here mean of distribution=μ=np=		160.00
and standard deviation σ=sqrt(np(1-p))=		11.31

probability =P(199.5<X<200.5)=P((199.5-160)/11.314)<Z<(200.5-160)/11.314)=P(3.49<Z<3.58)=0.9998-0.9997=0.0001