Respuesta :
Answer:
a) The size of the sample two dice so n= 2
b) The probability of getting a sum 10 on the two dice [tex]P(E) = \frac{3}{36} = \frac{1}{12}[/tex]
c) The probability of getting a sum of the dice at least 10 on the two dice so
[tex]P(E) =\frac{n(E)}{n(S)} = \frac{5}{36}[/tex]
d) The probability of getting that the sum of the dice is a prime number
[tex]P(E) =\frac{n(E)}{n(S)} = \frac{8}{36} = \frac{2}{9}[/tex]
Step-by-step explanation:
The total number of exhaustive cases throwing two dice(n(S) = 6 X6 =36
a) The size of the sample two dice so n= 2
b) Let E be the event of a getting a sum 10 on the two dice so the number of favorable to E = {(64),(4,6),(5,5) = 3
The probability of getting a sum 10 on the two dice
[tex]P(E) =\frac{n(E)}{n(S)} = \frac{3}{36} = \frac{1}{12}[/tex]
c) Let E be the event of a getting a sum of the dice at least 10 on the two dice so the number of favorable to E
At -least '10' means anything greater than or equal to ten
E = { (6,4),(4,6),(6,5),(5,6),(6,6)} =5
note:- (6,7),(7,6) events are not possible on dice because the die only '6' sides.
The probability of getting a sum of the dice at least 10 on the two dice so
[tex]P(E) =\frac{n(E)}{n(S)} = \frac{5}{36}[/tex]
d) Let E be the event of a getting a sum of the dice is a prime number on the two dice so the number of favorable to E
={(1,2),(2,1)(2,3),(3,2)(3,4),(4,3),(5,6),(6,5)} = 8
The probability of getting that the sum of the dice is a prime number
[tex]P(E) =\frac{n(E)}{n(S)} = \frac{8}{36} = \frac{2}{9}[/tex]
Conclusion:-
a) The size of the sample two dice so n= 2
b) The probability of getting a sum 10 on the two dice [tex]P(E) = \frac{3}{36} = \frac{1}{12}[/tex]
c) The probability of getting a sum of the dice at least 10 on the two dice
[tex]P(E) =\frac{n(E)}{n(S)} = \frac{5}{36}[/tex]
d) The probability of getting that the sum of the dice is a prime number
[tex]P(E) =\frac{n(E)}{n(S)} = \frac{8}{36} = \frac{2}{9}[/tex]
