# 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.

Last updated date: 26th Mar 2023

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Answer

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Hint- Here, a general formula for probability of occurrence of an event is used.

Given, we are tossing a die twice whose faces are marked with numbers 0,1,2,3,4,5.

As we know that the general formula for probability is given by

Probability of occurrence of an event$ = \dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of possible outcomes}}}}{\text{ }} \to {\text{(1)}}$

Here, the favourable event is that the product of digits on the upper face occurred when a die is rolled twice is zero. For this event to occur, zero appears at least once on the upper face of the die when it is rolled twice.

Here possible outcomes are $

\left( {0,0} \right),\left( {0,1} \right),\left( {0,2} \right),\left( {0,3} \right),\left( {0,4} \right),\left( {0,5} \right) \\

\left( {1,0} \right),\left( {1,1} \right),\left( {1,2} \right),\left( {1,3} \right),\left( {1,4} \right),\left( {1,5} \right) \\

\left( {2,0} \right),\left( {2,1} \right),\left( {2,2} \right),\left( {2,3} \right),\left( {2,4} \right),\left( {2,5} \right) \\

\left( {3,0} \right),\left( {3,1} \right),\left( {3,2} \right),\left( {3,3} \right),\left( {3,4} \right),\left( {3,5} \right) \\

\left( {4,0} \right),\left( {4,1} \right),\left( {4,2} \right),\left( {4,3} \right),\left( {4,4} \right),\left( {4,5} \right) \\

\left( {5,0} \right),\left( {5,1} \right),\left( {5,2} \right),\left( {5,3} \right),\left( {5,4} \right),\left( {5,5} \right) \\

$

Total number of possible outcomes$ = 36$

For product of the digits on the upper face when a die is rolled twice to be zero, the favourable outcomes are $\left( {0,0} \right),\left( {0,1} \right),\left( {0,2} \right),\left( {0,3} \right),\left( {0,4} \right),\left( {0,5} \right),\left( {1,0} \right),\left( {2,0} \right),\left( {3,0} \right),\left( {4,0} \right),\left( {5,0} \right)$.

Number of favourable outcomes$ = 11$

Using formula given by equation (1), we get

Required probability$ = \dfrac{{11}}{{36}}$.

Hence, the probability that the product of digits on the upper face is zero when a die is rolled twice is $\dfrac{{11}}{{36}}$.

Note- In these types of problems, the favourable event is considered as the event whose probability is asked. In this particular question for the product of digits on the upper face to be zero, either one of these digits should be zero or both of these digits should be zero.

Given, we are tossing a die twice whose faces are marked with numbers 0,1,2,3,4,5.

As we know that the general formula for probability is given by

Probability of occurrence of an event$ = \dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of possible outcomes}}}}{\text{ }} \to {\text{(1)}}$

Here, the favourable event is that the product of digits on the upper face occurred when a die is rolled twice is zero. For this event to occur, zero appears at least once on the upper face of the die when it is rolled twice.

Here possible outcomes are $

\left( {0,0} \right),\left( {0,1} \right),\left( {0,2} \right),\left( {0,3} \right),\left( {0,4} \right),\left( {0,5} \right) \\

\left( {1,0} \right),\left( {1,1} \right),\left( {1,2} \right),\left( {1,3} \right),\left( {1,4} \right),\left( {1,5} \right) \\

\left( {2,0} \right),\left( {2,1} \right),\left( {2,2} \right),\left( {2,3} \right),\left( {2,4} \right),\left( {2,5} \right) \\

\left( {3,0} \right),\left( {3,1} \right),\left( {3,2} \right),\left( {3,3} \right),\left( {3,4} \right),\left( {3,5} \right) \\

\left( {4,0} \right),\left( {4,1} \right),\left( {4,2} \right),\left( {4,3} \right),\left( {4,4} \right),\left( {4,5} \right) \\

\left( {5,0} \right),\left( {5,1} \right),\left( {5,2} \right),\left( {5,3} \right),\left( {5,4} \right),\left( {5,5} \right) \\

$

Total number of possible outcomes$ = 36$

For product of the digits on the upper face when a die is rolled twice to be zero, the favourable outcomes are $\left( {0,0} \right),\left( {0,1} \right),\left( {0,2} \right),\left( {0,3} \right),\left( {0,4} \right),\left( {0,5} \right),\left( {1,0} \right),\left( {2,0} \right),\left( {3,0} \right),\left( {4,0} \right),\left( {5,0} \right)$.

Number of favourable outcomes$ = 11$

Using formula given by equation (1), we get

Required probability$ = \dfrac{{11}}{{36}}$.

Hence, the probability that the product of digits on the upper face is zero when a die is rolled twice is $\dfrac{{11}}{{36}}$.

Note- In these types of problems, the favourable event is considered as the event whose probability is asked. In this particular question for the product of digits on the upper face to be zero, either one of these digits should be zero or both of these digits should be zero.

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