Question

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The Faces of A Die Bear Number 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.​

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Answer 1

Step-by-step explanation:

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)}

n(totaloutcomes)

n(gettingproductaszero)

=\dfrac{11}{36}

36

11

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

36

11

Remember

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

totalnumberofoutcomes

numberoffavourableoutcomes

Answer 2

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Faces of A Die Bear Number 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.

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S = {(0, 0), (0,1), (0,2), (1,0), (1,1), (1,2), (2,0), (2,1), (2,2), (3.0), (3,1), (3,2), (4.0), (4,1), (4,2), (5.0), (5,1), (5,2) }

∴ n(S) = 36 

Let A be the event that the product of digits 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 

∴ P(A) =  n ( A ) /n (S)

∴ P(A) = 11/36

∴ The probability that the product of the digits on the upper face is 11/36