Question

In tossing four fair dice, what is the probability of tossing, at most, one 3

Answer 1

Answer-

In tossing four fair dice, the probability of getting at most one 3 is 0.86.

Solution-

The probability of getting at most one 3 is, either getting zero 3 or only one 3.

$P(A) =The \ probability \ of \ getting \ zero \ 3 =(\frac{5}{6})(\frac{5}{6})(\frac{5}{6})(\frac{5}{6})=\frac{625}{1296} =0.48$                    ( ∵ xxxx )

$P(B) = The \ probability \ of \ getting \ only \ one \ 3 =(4)(\frac{1}{6})(\frac{5}{6})(\frac{5}{6})(\frac{5}{6}) = 0.38$                    ( ∵ 3xxx, x3xx, xx3x, xxx3 )

P(Atmost one 3) = P(A) + P(B) = 0.48 + 0.38 = 0.86