Question

In a single throw with two dice,find the probability that their sum is multiple either of 3 or 4

Answer 1

Answer:

Total no of sample=36

Multiple either 3 = 1 2, 2 1, 5 1, 4 2, 3 3, 2 4, 1 5, 6 3, 5 4, 4 5, 3 6, 6 6

= 12/36

= 1/3 Ans

Multiple either 4= 3 1, 2 2, 1 3, 6 2, 5 3, 4 4, 3 5, 2 6, 6 6

= 9/36

= 1/4 Ans

Answer 2

The probability that their sum is multiple either of 3 or 4 is $\frac{5}{9}$.

Step-by-step explanation:

Given:

In a single throw with two dice.

To Find:

The probability that their sum is multiple either of 3 or 4.

Their sum is multiple either of 3 or 4.

Solution:

As given,In a single throw with two dice.

Total  outcomes in a single throw with two dice

= (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).

It means, Total  numbers of outcomes in a single throw with two dice=36

As given,their sum is multiple either of 3 or 4.

Then, Favorable Outcomes=(1,2),(1,3),(1,5),(2,1),(2,2),(2,4),(2,6),(3,1),(3,3)

            ,(3,5),(3,6),(4,2),(4,4),(4,5),(5,1),(5,3),(5,4),(6,2),(6,3),(6,6).

It means, The numbers of favorable Outcomes =20.

The probability that their sum is multiple either of 3 or 4

                                                 $=\frac{The\ numbers\ of\ favorable\ Outcomes}{Total\ outcomes\ in\ a\ single\ throw\with\ two\ dice }$

                                                $=\frac{20}{36} =\frac{5}{9}$

Thus,the probability that their sum is multiple either of 3 or 4 is $\frac{5}{9}$.

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