PLEASE HELP AND ROUND If a fair die is rolled 7 times, what is the probability, rounded to the nearest thousandth, of getting at most 2 threes?

Question

If a fair die is rolled 7 times, what is the probability, rounded to the nearest thousandth, of getting at most 2 threes?

Answer 1

Pr(< two threes)

= Pr(0,1 or 2 threes)

= 7C0(1/6)^0(5/6)^7 + 7C1(1/6)^1(5/6)^6 + 7C2(1/6)^2(5/6)^5

=7!/7!0!(1)(5/6)^7 +7!/1!6!(5^6)/6^7 + 7!/2!5!(5^5)/(6^7)

= 5^7/6^7 +7(5^6)/6^7 + 21(5^5)/6^7

=(5^5/6^7)[5^2 +7(5) +21]

=3125/279936)(81)

=3125/3456

=.904224537

= 90.422% probability of at most two threes in seven fair die rolls

1/6 chance for a 3 on each roll

5/6 chance of no 3 on each roll

Probability of zero 3's in 7 rolls = (5/6)^7

Probability of exactly one 3 in 7 rolls = 7x(1/6)(5/6)^6

Probability of exactly two 3's in 7 rolls = 21(1/6)^2(5/6)^5

sum them to get Probability of no more than two 3's in 7 rolls

Pr(7 3's) = 7C7(1/6)^7 = 1/6^7= .00004 = 0.000%

Pr(6 3's) = 7C6(1/6)^6(5/6) = 7(5)/6^7 =35/6^7 = .0135%

Pr(5 3's) = 7C5(1/6)^5(5/6)^2= 21(5^2)/6^7 = 525/6^7=.188%

Pr(4 3's) = 7C4(1/6)^4(5/6)^3= 35(5^3)/6^7= 4375/6^7= .2605%

Pr(3 3's) = 7C3(1/6)^3(5/6)^4 = 35(5^4)/6^7)= 21875/6^7 = 7.814%