Question

4) (10 pts) You are doubtful about the fairness of a six-side die. It's expected that 50% of the time you should get the numbers 1, 2, or 3 a) After 20 rolls, what is the standard deviation you'd expect of the number of times you get 1, 2, or 3? b) After 20 tosses you find that you got these numbers only 40% of the time. Do you make an accusation? many rolls would it take for you to conclude that a continued 40% average c) How indicates loaded dice?

Answer 1

Solution

Let X = Number of times 1, 2 or 3 turns up when a die is rolled n times. Then, X ~ B(n, p), where

p = probability of the die turning up 1, 2 or 3 = 0.5 for a fair die. ………………………………. (1)

Back-up Theory

If X ~ B(n, p). i.e., X has Binomial Distribution with parameters n and p, where n = number of trials and p = probability of one success, then probability mass function (pmf) of X is given by

p(x) = P(X = x) = (ⁿC_x)(p^x)(1 - p)^{n – x}, x = 0, 1, 2, ……. , n …………………..(2)

[The above probability can also be directly obtained using Excel Function: Statistical, BINOMDIST……………………………………….(1a)

Mean (average) of X = E(X) = µ = np………………………………………………..(3)

Variance of X = V(X) = σ² = np(1 – p)…………………………………………………..(4)

Standard Deviation of X = SD(X) = σ = √{np(1 – p)} ………………………………………...(5)

Now to work out the solution,

Part (a)

Given n = 20 and p = 0.5, vide (1) and (5),

Expected standard deviation of the number of times 1, 2 or 3 turns up = √(20 x 0.5 0.5)

= 2.2361 Answer

Part (b)

40% of 20 = 8.

We will first find the probability of getting 1, 2 or 3 eight times when the die is rolled 20 times.

i.e., P(X = 8) = 0.1201 [vide (1a)]

As a general practice, 5% i.e., 0.05 is considered a low probability. In view of this, the above probability is quite high implying thereby that it is only likely that 20 rolls would result in 8 times of

1, 2 or 3. Thus, there is no room for accusation.   Answer

Part (c)

Again, considering 0.05 as low probability, here we need to find n such that P(X = 0.4n) < 0.5.

From the results of Part (b), it is clear that n must greater than 20.

The following table gives, P = P(X = 0.4n) for various values n and p = 0.5, which has found using (1a).

p = 0.5
n	n x 0.4	P
25	10	0.0974166
30	12	0.0805531
40	16	0.0571637
45	18	0.0487684
42	17	0.0579034
44	18	0.058522

From the above table, it is clear that n = 45 Answer

DONE

[Going beyond, answer to Part (c) could also be obtained by using Normal approximation of Binomial probabilities]