there is a six sided dice rolled 40 times what is the expiremental and theoretical probability of rolling a 6, 1 or 6, no more then 4, greater then 2.

Question

number of dice frequency

1            2

2            10

3             5

4             4

5            7

6            12

Answer 1

all the probabilities, theoretical or experimental are nearly certain

rounded off to nearest percent = 100%

experimental probability = 12/40 =.3 = 30% chance of a 6 in one roll

14/40 = 7/20 =3.5/10 = 35% chance of 1 or 6 in one roll

21/40 = 10.5/20= 5.25/10 = 52.5% chance of no more than 4

28/40 = 70% chance of more than 2

in 40 rolls

P(6) = 1-(.7)^40 = 1-.0000013 = 99.99913% for a 6 in 40 rolls

P(1 or 6) = 1-.65^40 = 1- .000000033 = 99.9999933% for 1 or 6 in 40 rolls

P(<5) = 1-.485^2 = 1- .2.7x10^-13 = 99.999999999913% for no more than 4 in 40 rolls

P(>2) = 1-.3^40 = 1- 1.2x10^-21 =99.9999999999999999999% for more than 2 in 40 rolls

theoretical probabilities are

Pr(6) = 1/6=about 16.7% in one roll

Pr(6) =1-(5/6)^2 = 11/36 = about 30.6% in two rolls

P(6) =1-(5/6)^3 = 91/216 = about 42.1% in 3 rolls

Pr(6) =1=(5/6)^40 in 40 rolls

=1- 5^40/6^40

=1 - 9.0949947018x10^27/1.336749454x10^31

=1 - .0006803814039

=99.93% = slightly worse than the experimental probability

P(1 or 6) = 1-(2/3)^40= slightly worse than the experimental probability

P(<5) = 1-(1/3)^40= slightly better than the experimental probability

P(>2) = 1-(1/3)^40 = slightly worse than the experimental probability

the close similarity in the odds between experimental and theoretical probability suggests the dice are fair, not loaded.

P(exactly one 6) = 40C6(1/6)^6(5/6)^34

=40!/6!34!(5^34/6^40)

=40(39)(38)(37)(36)(35)/6(5)(4)(3)(2)(5.820766091x10^23/1.336749454x10^31)

=3.838380x10x^6 x 4.354418155x10^-8

=1.671391156 x 10^-1

=about 0.167 = 16.7% chance of exactly one 6 in 40 rolls