Respuesta :
Answer:
[tex]P(E_1 or E_2) = \frac{7}{12}[/tex]
Step-by-step explanation:
Given
Two cubes of side 1 - 6
Required
Probability that the outcome of the roll is an odd sum or a sum that is a multiple of 5
First, the sample space needs to be listed;
Let [tex]C_r[/tex] represent the Red cube
[tex]C_b[/tex] represent the Blue cube
S represent the sample space
[tex]C_r = (1,2,3,4,5,6)\\C_b = (1,2,3,4,5,6)\\S = (2,3,4,5,6,7,3,4,5,6,7,8,4,5,6,7,8,9,5,6,7,8,9,10,6,7,8,9,10,11,7,8,9,10,11,12)[/tex]S is gotten by getting the sum of [tex]C_r[/tex] and [tex]C_b[/tex]
[tex]n(S) = 36[/tex]
Calculating the Probability
Let [tex]E_1[/tex] represent the event that an outcome is an odd sum
[tex]E_1 = (3,5,7,3,5,7,5,7,9,5,7,9,7,9,11,7,9,11)[/tex]
[tex]n(E_1) = 18[/tex]
[tex]P(E_1) = \frac{n(E_1)}{n(S)}[/tex]
[tex]P(E_1) = \frac{18}{36}[/tex]
Let [tex]E_2[/tex] represent the event that an outcome is a multiple of 5
[tex]E_2 = (5,5,5,5,10,10,10)[/tex]
[tex]n(E_2) = 7[/tex]
[tex]P(E_2) = \frac{n(E_2)}{n(S)}[/tex]
[tex]P(E_2) = \frac{7}{36}[/tex]
Let [tex]E_3[/tex] represent the event that an outcome is an odd sum and a multiple of 5
[tex]E_3 = E_1 and E_2[/tex]
[tex]E_3 = (5,5,5,5)[/tex]
[tex]n(E_3) = 4[/tex]
[tex]P(E_3) = \frac{n(E_3)}{n(S)}[/tex]
[tex]P(E_3) = \frac{4}{36}[/tex]
Calculating [tex]P(E_1 or E_2)[/tex]
[tex]P(E_1 or E_2) = P(E_1) + P(E_2) - P(E_1 and E_2)[/tex]
[tex]P(E_1 or E_2) = P(E_1) + P(E_2) - P(E_3)[/tex]
[tex]P(E_1 or E_2) = \frac{18}{36} + \frac{7}{36} - \frac{4}{36}[/tex]
[tex]P(E_1 or E_2) = \frac{18 + 7 - 4}{36}[/tex]
[tex]P(E_1 or E_2) = \frac{21}{36}[/tex]
[tex]P(E_1 or E_2) = \frac{7}{12}[/tex]
Hence, the probability that the outcome of the roll is an odd sum or a sum that is a multiple of 5 is [tex]\frac{7}{12}[/tex]
