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Hint- Here, we will proceed by writing down all the possible cases occurred when two dice are thrown simultaneously and then will find out those cases that will give the product of the numbers appearing as a prime number and then use the general formula for probability.

“Complete step-by-step answer:”

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}}}}$

When two dice are thrown at the same time there are a total 36 possible outcomes which are given as under.

(1,1),(1,2) ,(1,3) ,(1,4) ,(1,5) ,(1,6), (2,1) ,(2,2) ,(2,3) ,(2,4) ,(2,5) ,(2,6) ,(3,1) ,(3,2) ,(3,3) ,(3,4) ,(3,5) ,(3,6) ,(4,1) ,(4,2) ,(4,3) ,(4,4) ,(4,5) ,(4,6) ,(5,1) ,(5,2) ,(5,3) ,(5,4) ,(5,5) ,(5,6) ,(6,1) ,(6,2) ,(6,3) ,(6,4) ,(6,5) ,(6,6) where we have (a,b) in which a represents the number appearing on the first dice and b represents the number appearing on the second dice when two dice are thrown simultaneously.

Total number of possible outcomes = 36

Here, the favourable event is obtaining a prime number as the product of the numbers appearing on the two dice which are thrown simultaneously.

As prime numbers are 2,3,5,7,11,13,17,19,23,29,31,etc

Favourable outcomes are (1,2),(1,3),(1,5),(2,1),(3,1),(5,1).

Only these six cases are giving the product of the numbers appearing on the two dice as a prime number.

Number of favourable outcomes = 6

Therefore, probability that the product is a prime number$ = \dfrac{{\text{6}}}{{{\text{36}}}} = \dfrac{1}{6}$.

Hence, the required probability is $\dfrac{1}{6}$.

Note- In this particular problem, possible cases are (1,2),(1,3),(1,5),(2,1),(3,1),(5,1) because (1,2) gives the product as 2, (1,3) gives the product as 3, (1,5) gives the product as 5, (2,1) gives the product as 2, (3,1) gives the product as 3, (5,1) gives the product as 5 where all the numbers 2,3 and 5 are prime numbers.

“Complete step-by-step answer:”

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}}}}$

When two dice are thrown at the same time there are a total 36 possible outcomes which are given as under.

(1,1),(1,2) ,(1,3) ,(1,4) ,(1,5) ,(1,6), (2,1) ,(2,2) ,(2,3) ,(2,4) ,(2,5) ,(2,6) ,(3,1) ,(3,2) ,(3,3) ,(3,4) ,(3,5) ,(3,6) ,(4,1) ,(4,2) ,(4,3) ,(4,4) ,(4,5) ,(4,6) ,(5,1) ,(5,2) ,(5,3) ,(5,4) ,(5,5) ,(5,6) ,(6,1) ,(6,2) ,(6,3) ,(6,4) ,(6,5) ,(6,6) where we have (a,b) in which a represents the number appearing on the first dice and b represents the number appearing on the second dice when two dice are thrown simultaneously.

Total number of possible outcomes = 36

Here, the favourable event is obtaining a prime number as the product of the numbers appearing on the two dice which are thrown simultaneously.

As prime numbers are 2,3,5,7,11,13,17,19,23,29,31,etc

Favourable outcomes are (1,2),(1,3),(1,5),(2,1),(3,1),(5,1).

Only these six cases are giving the product of the numbers appearing on the two dice as a prime number.

Number of favourable outcomes = 6

Therefore, probability that the product is a prime number$ = \dfrac{{\text{6}}}{{{\text{36}}}} = \dfrac{1}{6}$.

Hence, the required probability is $\dfrac{1}{6}$.

Note- In this particular problem, possible cases are (1,2),(1,3),(1,5),(2,1),(3,1),(5,1) because (1,2) gives the product as 2, (1,3) gives the product as 3, (1,5) gives the product as 5, (2,1) gives the product as 2, (3,1) gives the product as 3, (5,1) gives the product as 5 where all the numbers 2,3 and 5 are prime numbers.

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