Question

The faces of a die bear numbers 0,1,2,3,4,5. If the die is rolled twice, then find the probability that the product of digits on the upper face is zero. In detail please

Answer 1

Answer:

S= { (0,0) (0,1),(0,2),(0,3),(0,4),(0,5),(1,0),(1,1),(1,2),(1,3),(1,4),(1,5),(2,0),(2,1),(2,2),(2,3),(2,4),(2,5),(3,0),(3,1),(3,2),(3,3),(3,4),(3,5),(4,0),(4,1),(4,2),(4,3),(4,4),(4,5),(5,0),(5,1),(5,2),(5,3),(5,4),(5,5) }

n(S) = 36

Now , let A is the event that the product of digit on the upper face is zero

A ={(0,0) (0,1),(0,2),(0,3),(0,4),(0,5),(1,0) ,(2,0),(3,0),(4,0),(5,0) }. n(A) = 11

Now , p(A) = n(A) / (S) = 11 / 36

P(A) = 11 / 36

Answer 2

sample space ={ 0×0,0×1,0×2,0×3,0×4,0×5,1×0,1×1,1×2,1×3,1×4,1×5,2×0,2×1,2×2,2×3,2×4,2×5,3×0,3×1,3×2,3×3,3×4,3×5,4×0,4×1,4×2,4×3,4×4,4×5,5×0,5×1,5×2,5×3,5×4,5×5.}

n (S)=30

we have to get product 0 on upper side

therefore A={0×1,0×2,0×3,0×4,0×5,1×0,2×0,3×0,4×0,5×0}

n (A)=10

probability =

p (A) =n(A)÷n(S)

= 10 ÷ 30

= 1/3