ANSWER ASAPPPP FORR BRAINLIEST
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Answer:
<em>The probability that the sum is odd or a multiple of 5</em>
<em>P(E₁∪E₂) = 0.58 = 58%</em>
Step-by-step explanation:
<u><em>Step ( i ) :- </em></u>
<em>Given the two dice are thrown ,The total number of exhaustive cases </em>
<em>n(S) = 6² = 36</em>
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
<u><em>Step(ii):-</em></u>
The probability that the event of getting the sum is odd on two dice
The probability that the event of getting the sum is multiple of '5' on two dice
The probability that the event of getting the sum is odd and multiple of '5' on two dice
<em>The probability that the sum is odd or a multiple of 5</em>
<em>P(E₁∪E₂) = P(E₁) + p(E₂) - P(E₁ ∩ E₂)</em>
<em> = </em><em></em>
<em> </em><em></em>
<em>P(E₁∪E₂) = 0.58 = 58%</em>
<u><em>Final answer</em></u><u>:-</u>
<em>The probability that the sum is odd or a multiple of 5</em>
<em>P(E₁∪E₂) = 0.58 = 58%</em>
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