Question

If we were to toss a single die (singular for dice) 1000 times, how many 6’s could we expect over the long run? What is the standard deviation?

Answer 1

Consider n as the number of times the die is thrown, p as the probability of occurring 6 in a single throw and q as the probability of not occurring a 6 in a single throw. Now, to find the expected number of 6’s that may appear use the binomial distribution formula given as the expected number of occurrences of 6 = $np$. Now, find the variance of the event by using the binomial distribution formula given as ${{\sigma }^{2}}=npq$ where ${{\sigma }^{2}}$ denotes the variance. Take the square root of the variance to get the standard deviation.

Complete step-by-step solution:
Here we have been given the number of time a die is thrown and we are asked to find the expected number of times 6 may occur. Also we have to find the standard deviation. Here we need to apply the binomial distribution formula to get the answer.
Now, let us consider the number of times the die is thrown is n, probability of occurring 6 in a single throw as p and probability of not occurring a 6 in a single throw as q. We know that a die is cubical in shape and the outcomes can be 1, 2, 3, 4, 5 or 6. Clearly we can see that we have only one favourable outcome (6) and the number of possible outcomes is 5 (1, 2, 3, 4, 5). Therefore we get,
$\Rightarrow n=1000,p=\dfrac{1}{6}$ and $q=\dfrac{5}{6}$
(i) Now, the expected number of times of the occurrence of a variable in binomial distribution is given by the product of n and p. so we get,
$\Rightarrow$ Expected number of occurrence of 6 = $1000\times \dfrac{1}{6}$
$\therefore$ Expected number of occurrence of 6 = $\dfrac{500}{3}$
(ii) To find the standard deviation first we need to calculate the variance because the square root of a variance is the standard deviation. In binomial probability distribution the variance is given by the formula ${{\sigma }^{2}}=npq$ where ${{\sigma }^{2}}$ denotes the variance. So we get,
$\begin{aligned} & \Rightarrow {{\sigma }^{2}}=1000\times \dfrac{1}{6}\times \dfrac{5}{6} \\\ & \Rightarrow {{\sigma }^{2}}=\dfrac{5000}{36} \\\ \end{aligned}$
Taking square root both the sides we get,
$\begin{aligned} & \Rightarrow \sigma =\sqrt{\dfrac{5000}{36}} \\\ & \Rightarrow \sigma =\dfrac{50\sqrt{2}}{6} \\\ & \therefore \sigma =\dfrac{25\sqrt{2}}{3} \\\ \end{aligned}$

Note: Remember the formulas used in the binomial probability distribution. If x is the desired number of success of an event containing n trials then the probability of the success is given as $P\left( x \right){{=}^{n}}{{C}_{x}}{{p}^{x}}{{q}^{1-x}}$. This formula is used in the binomial probability distribution. These formulas are used in calculation of certain terms used in statistics.