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:The probability that the number 6 will appear between 150 and 200 times inclusively is 0.9258 The probability that the number 5 will appear less than 150 times is 0.1762.Solution:Given, One thousand independent rolls of a fair die will be made. First, let’s compute an approximation to the probability that number 6 will appear between 150 and 200 times.  The die is rolled 1,000 times. These are independent trials which are Bernoulli trials since a die shows either a 6 or not.  Thus, n = 1000. The probability of success is constant, p = 1/6. We can use the binomial model.  Let X be the number of times the die shows a 6.  We can use the normal approximation to the binomial model  Because np = 1000 • 1/6 > 10 and n(1 − p) = 1000 • 5 6 > 10. x ∼ Bin(166.7, 11.79). But, using the normal approximation, let Y be normally distributed using the parameters from the binomial model. That is, in the normal model [tex]\mu=\mathrm{np} \text { and } \sigma=\sqrt{\mathrm{np}(1-\mathrm{p})}[/tex]So, let y ∼ N(166.7, 11.79). There is one small complication. We are using a continuous distribution to model a discrete distribution. We must correct for this discrepancy using the continuity correction which is bolded in the equation below.  P(150 ≤ Y ≤ 200) = P(Y ≤ 200) − P(Y ≤ 150)  [tex]\begin{array}{l}{=\mathrm{P}\left(Z \leq \frac{200+0.5-166.7}{11.79}\right)-\mathrm{P}\left(Z \leq \frac{150-0.5-166.7}{11.79}\right)} \\\\ {=\Phi(2.87)-\Phi(-1.46)} \\\\ {=0.998-0.072} \\\\ {=0.9258}\end{array}[/tex]Now, we consider the probability that we obtain less than 150 6s if exactly 200 5s have been shown. This is a conditional probability problem. Let Y be the number of 6s that appear and let X be the number of 5s that appear.  Then we need to compute P(Y < 150|X = 200).  That is, we consider that 200 of the rolls show a 5, therefore, we have 800 trials remaining that can display anything except a 5, so n = 800 now.  Since we have already considered those 200 trials where a 5 appears, the probability of success (a 6 appears) is p = 1/5.  Again, we use the normal approximation due to the large value for n and since the success/failure conditions hold (np > 10 and n(1 − p) > 10).  [tex]\text { We know that } \mu=\mathrm{np}=800 \times \frac{1}{5}=160 \text { and } \sigma=\sqrt{800 \times \frac{1}{5} \times \frac{4}{5}}=11.31[/tex]Let Y [tex]\times[/tex] ∼ N(160, 11.31). [tex]\text { Then, } \mathrm{P}(\mathrm{Y} \times <150)=\mathrm{P}\left(Z<\frac{150-0.5-160}{11.31}\right)=\Phi(-0.93)=0.1762[/tex]Hence, the probabilities are 0.9258 and 0.1762.