Question

10. A pair of dice is rolled 180 times. What is the probability that a total of six occurs
a) Atleast 30 times b)between 33 and 42 times inclusive. Use Normal approximation to Binomial distribution

Answer 1

Answer on Question #45191 – Math – Statistics and Probability

Question. A pair of dice is rolled 180 times. What is the probability that a total of six occurs

a) At least 30 times;

b) between 33 and 42 times inclusive.

Use Normal approximation to Binomial distribution.

Solution. We can assume that one dice is rolled 360 times. Normal approximation of Binomial distribution: $P(n, k_1, k_2) \approx \Phi\left(\frac{k_2 - np}{\sqrt{npq}}\right) - \Phi\left(\frac{k_1 - np}{\sqrt{npq}}\right)$ , where $\Phi(x) = \frac{1}{\sqrt{2\pi}} \int_{0}^{x} e^{-\frac{t^2}{2}} dt$ is the function of Laplas (this function is tabulated).

a) In this case $n = 360, k_1 = 30, k_2 = 360, p = \frac{1}{6}, q = \frac{5}{6} \cdot \frac{k_2 - np}{\sqrt{npq}} \approx 42.43, \frac{k_1 - np}{\sqrt{npq}} \approx -4.24$ .

$P(360, 30, 360) \approx \Phi(42.43) - \Phi(-4.24) = \Phi(42.43) + \Phi(4.24) \approx 0.5 + 0.49999 = 0.99999.$

b) In this case $n = 360, k_1 = 33, k_2 = 42, p = \frac{1}{6}, q = \frac{5}{6} \cdot \frac{k_2 - np}{\sqrt{npq}} \approx -2.55, \frac{k_1 - np}{\sqrt{npq}} \approx -3.82$ .

$P(360, 33, 42) \approx \Phi(-2.55) - \Phi(-3.82) = \Phi(3.82) - \Phi(2.55) \approx 0.49993 - 0.49446 = 0.00547.$

Answer.

a) $P(360, 30, 360) \approx 0.99999$ .

b) $P(360, 33, 42) \approx 0.00547$ .

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