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
[tex]P(E_{1}) = \frac{n(E)}{n(S)} = \frac{18}{36}[/tex]
The probability that the event of getting the sum is multiple of '5' on two dice
[tex]P(E_{2}) = \frac{n(E_{2} )}{n(S)} = \frac{7}{36}[/tex]
The probability that the event of getting the sum is odd and multiple of '5' on two dice
[tex]P(E_{1} n E_{2}) = \frac{n(E_{1} n E_{2} )}{n(S)} = \frac{4}{36}[/tex]
The probability that the sum is odd or a multiple of 5
P(E₁∪E₂) = P(E₁) + p(E₂) - P(E₁ ∩ E₂)
= [tex]\frac{18}{36} + \frac{7}{36} - \frac{4}{36}[/tex]
[tex]\frac{18+7}{36} - \frac{4}{36} = \frac{25-4}{36} = \frac{21}{36}[/tex]
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%
The probability that the sum is odd or a multiple of 5 is 56 %.
When two dice are rolled, then
Total number of outcomes is [tex]=6^{2}=36[/tex]
Outcomes for the sum is odd or a multiple of 5 ,
[tex](1,2),(2,1),(2,3),(3,2),(1,4),(4,1),(1,6),(6,1),(2,5),(5,2),(3,4)\\\\(4,3),(3,6),(6,3),(4,5),(5,4),(6,5),(5,6),(4,6),(6,4)[/tex]
Number of favourable outcomes are [tex]=20[/tex]
the probability that the sum is odd or a multiple of 5 is,
[tex]P(E)=\frac{20}{36} =0.56[/tex]
Learn more about the probability here:
https://brainly.com/question/24756209
