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

Answer 1

probability of products
0*0
0*1
0*2
0*3
0*4
0*5
1*0
1*1....
so we can make out that events would be 6*6=36
and no of times 0 will be the product is
6+5=11
( six times when 0 is multiplied by another no
and rest 5 times when another no is multiplied by 0)
so we have probability=11/36

Answer 2

Answer

Here we have>

No. of possible outcomes(s) =>

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(total outcomes) = 36

Now, let the number of favorable outcomes i.e the product of two numbers are zero be =>

n(getting product as zero)

=(0,0) (0,1),(0,2),(0,3),(0,4),(0,5),(1,0) ,(2,0),(3,0),(4,0),(5,0).

i.e n(getting product as zero) = 11.

Hence the probability that the product of digits on the upper face is zero will be

P(getting the product as zero)

=$\dfrac{n(getting product as zero)}{n(total outcomes)}$

=$\dfrac{11}{36}$

Hence the probability of getting the product of digits on the upper face is zero is$\dfrac{11}{36}$

Remember

Probability =$\dfrac{number of favourable outcomes}{total number of outcomes}$