If a fair die is rolled 7 times, what is the probability, rounded to the nearest thousandth, of getting at least 6 threes?

Answer 1

fair 6 sided die rolled 7 times

odds of at least 6 three's

= probability of 7 three's plus probability of 6 three's

= Pr(7 3's) + Pr(6 3's)

= (1/6)^7 + 7C6(1/6)^6(5/6)^1

= 1/6^7 + (7!/6!)(1/6^6)(5/6)

= 1/6^7 + 35/6^7)

=1/279936 +35/279,936

= 36/279,936

= 0.000128600823

rounded to nearest thousandths

=0.000

= zero point zero zero zero

= zero, but still possible, however unlikely