Question

You roll two six-sided dice. What is the probability that the sum is odd or a multiple of 5? Round your answer to the nearest tenth of a percent.

Answer 1

Answer:

The probability that the sum is odd or a multiple of 5

P(E₁∪E₂) = 0.58 = 58%

Step-by-step explanation:

Step ( i ) :-    

Given the two dice are thrown ,The total number of  exhaustive cases

n(S) =  6² = 36

Let 'E₁' be the event of getting the sum is odd on two dice

E₁ = { (1,2)(1,4),(1,6),(2,1),(2,3),(2,5),(3,2)(3,4),(3,6),(4,1)(4,3),(4,5),(5,2),(5,4),(5,6),(6,1),(6,3),(6,5)}

n(E₁) = 18

Let 'E₂' be the event of getting the sum is multiple of 5 on two dice

E₂ = { (1,4),(2,3), (3,2),(4,1),(4,6),(5,5),(6,4)}  

n(E₂) = 7  

n(E₁∩E₂) = {(1,4),(2,3),(3,2),(4,1)} = 4

Step(ii):-

The probability that the event of getting the sum is odd on two dice

$P(E_{1}) = \frac{n(E)}{n(S)} = \frac{18}{36}$

The probability that the event of getting the sum is multiple of '5' on two dice

$P(E_{2}) = \frac{n(E_{2} )}{n(S)} = \frac{7}{36}$

The probability that the event of getting the sum is odd and multiple of '5' on two dice

$P(E_{1} n E_{2}) = \frac{n(E_{1} n E_{2} )}{n(S)} = \frac{4}{36}$

The probability that the sum is odd or a multiple of 5

P(E₁∪E₂) = P(E₁) + p(E₂) - P(E₁ ∩ E₂)

              =    $\frac{18}{36} + \frac{7}{36} - \frac{4}{36}$

              $\frac{18+7}{36} - \frac{4}{36} = \frac{25-4}{36} = \frac{21}{36}$

P(E₁∪E₂) = 0.58 = 58%

Final answer:-

The probability that the sum is odd or a multiple of 5

P(E₁∪E₂) = 0.58 = 58%