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. Solve the word problem

Answer 1

I think 11/36. please verify and inform me too. it's a good question.
I solved in negative sense first found not to get zero so it will be 5/6*5/6 = 25/36
so rest is for zero so 11/36

Answer 2

Answer:

The probability that the product of upper face is zero, will be $\dfrac{11}{36}$

Step-by-step explanation:

Given,

the numbers marked on the faces of die are

0, 1, 2, 3, 4, 5, 6

Since, we are rolling the dice twice, So, the possible outcomes are

(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)

The outcomes whose product will be zero are

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

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

Hence, we can see that total number of outcomes = 36

and total number of outcomes whose product will be zero = 11

So, the probability that the product of digit on upper face will zero is

$P\ =\ \dfrac{number\ of\ outcomes\ whose\ product\ is\ zero}{total\ number\ of\ outcome}$

   $=\dfrac{11}{36}$

So, the probability that the product of upper face is zero, will be $\dfrac{11}{36}$