Question

2 dice are thrown simultaneously. What is the probability that the sum of the numbers on the faces is divisible by either 3 or 5?

Answer 1

Answer:

here is the ans

Explanation:

The sample space S when two dice are thrown together is:

(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).

Therefore, n(S)=36.

Let A denote the event that sum of number is not divisible by 4 or 5 that is:  

(1,1),(1,2),(1,5),(1,6),

(2,1),(2,4),(2,5),

(3,3),(3,4),(3,6),

(4,2),(4,3),(4,5),

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

(6,1),(6,3),(6,5)

n(A)=20, therefore, probability that sum of number is not divisible by 4 or 5 is:

P(A)=   n(S)/ n(A)  =  20/36   = 5/9

Hence, probability that sum of number on the faces is neither divisible by 4 nor by 5 is   5/9...  

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Answer 2

Given:

Two dice are thrown simultaneously.

To Find:

The probability that the sum of the numbers on the faces is divisible by either 3 or 5.

Solution:

The given problem can be solved by using the concepts of probability.

1. It is given that two dice are thrown at random. Hence the total number of possibilities will be 6 x 6 i.e 36. The following are the possible outcomes.

• (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)

2. Probability of an event is defined as the favorable outcomes divided by the total number of outcomes.

=> P(event happening) = (favorable outcomes)/(total outcomes).

3. Probability ( sum of two faces is divisible by 3 or 5 ) = Probability (the sum divisible by 5) + Probability (the sum divisible by 5) - Probability (the sum divisible by both 3 and 5)

4. Probability ( sum divisible by 5 ) = (total numbers whose sum is divisible by 5) / (total number of outcomes).

=> Probability ( sum divisible by 5 ) = 7/36.

5. Probability ( sum divisible by 3 ) = (total numbers whose sum is divisible by 3) / (total number of outcomes).

=> Probability ( sum divisible by 3 ) = 12/36.

6. Probability ( sum divisible by both 3 and 5 ) = (total numbers whose sum is divisible by both 3 and 5) / (total number of outcomes).

=> Probability ( sum divisible by both 3 and 5 ) = 0/36.

7. Probability ( sum divisible by either 3 or 5 ) = (total numbers whose sum is divisible by either 3 or 5) / (total number of outcomes).

=> Probability ( sum divisible by either 3 or 5 ) = 19/36.

Therefore, the probability that the sum of the numbers is divisible by either 3 or 5 is 19/36.