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.

Short answer

The probability that the number$5$will appear less than$150$times is$0.1762.$

Step-by-step solution

Step 1. Given Information.

The die is rolled $1000$times.

The trials are independent and they tend to follow Bernoulli trials.

The probability of success is constant$p=\frac{1}{6}$.

Step 2. Find the probability that the number 5 will appear less than 150 times if the number 6 appears exactly 200 times.

Let $X$be the number of times the die shows a $6$.

Now let us calculate the probability that $6$will appear between $150and200$times inclusively.

$$\begin{gathered} \mu =np \\ =1000\times \frac{1}{6} \\ =166.6667 \end{gathered}$$

$$\begin{gathered} \sigma =\sqrt{np\left(1-p\right)} \\ =11.7851 \end{gathered}$$

$$\begin{gathered} Also,np=1000\times \frac{1}{6} \\ =166.667>10 \\ and,nq=n\left(1-p\right) \\ =1000\times \frac{5}{6} \\ =833.333>10 \end{gathered}$$

Since $np>10andnq>10$use the normal approximation to the binomial model.

Now let $Y$be normally distributed using the parameters from the binomial model.

localid="1646745007204" $$\begin{gathered} Y~N\left(166.6667,11.7851\right) \\ P\left(150\le X\le 200\right) \\ =P\left(\frac{150-\mu }{\sigma }\le \frac{X-\mu }{\sigma }\le \frac{200-\mu }{\sigma }\right) \\ =P\left(\frac{150-0.5}{11.785}\le Z\le \frac{200-0.5}{11.7851}\right) \\ ByContinuityCorrection \\ =P\left(\frac{150-0.5-166.6667}{11.785}\le Z\le \frac{200+0.5-166.6667}{11.7851}\right) \\ =P\left(\frac{149.5-166.6667}{11.785}\le Z\le \frac{200.5-166.6667}{11.7851}\right) \\ =P\left(-1.4566\le Z\le 2.87085\right) \\ =P\left(-1.46\le Z\le 2.87\right) \\ =\phi \left(2.87\right)-\phi \left(-1.46\right) \\ =0.998-0.072 \\ =0.9258 \end{gathered}$$

Hence, the probability that $6$will appear between $150and200$times inclusively is $0.9258.$

Step 3. Consider the probability to obtain less than 150 6's if exactly 200 5's have been shown.

This is a conditional probability: $P\left(Y<150|X=200\right)$

In the remaining $800$trials there are role="math" localid="1646745883182" $5$choices $\left\{1,2,3,4,5\right\}$when the die is rolled.

So , in this case the probability of getting upper face on die is $\frac{1}{5}$.

Let $X$denote number of times $5$(upper face) occurs when a die is rolled$800$times.

$$\begin{gathered} Y~Binom\left(n=800,p=\frac{1}{5}\right) \\ P\left(Y<150|n=200,p=\frac{1}{5}\right) \end{gathered}$$

Here,

Now using normal approximation let us find the required probability.

The mean and standard deviation will be:

$$\begin{gathered} \mu =np=800\times \frac{1}{5} \\ =160 \\ \sigma =\sqrt{np(1-p)} \\ =\sqrt{800\times \frac{1}{5}\times \frac{4}{5}} \\ =11.31 \end{gathered}$$

Step 4. Compute the probability that the number 5 will appear less than 150 times.

$$\begin{gathered} P\left(Y<150|n=200,p=\frac{1}{5}\right) \\ =P\left(Y<150\right) \\ =P\left(Z<\frac{150-0.5-160}{11.31}\right) \\ Continuitycorrection \\ =P\left(Z<-0.92838\right) \\ =\phi \left(-0.93\right) \\ =0.1762 \end{gathered}$$

Therefore, the probability that the number $5$ will appear less than $150$ times.