Question

(1 point) Library/Mizzou/Finite_Math/Probability_Introduction/Prob1.Pg Suppose that you roll two 6 sided dice. a) What is the size of the sample space? b) What is the probability that the sum of the dice is a 10? c) What is the probability that the sum of the dice is at least a 10? d) What is the probability that the sum of the dice is a prime number?

Answer 1

Answer:

a) The size of the sample two dice so n= 2

b) The probability of getting a sum 10 on the two dice $P(E) = \frac{3}{36} = \frac{1}{12}$

c)  The probability of getting a sum of the dice at least 10 on the two dice so

$P(E) =\frac{n(E)}{n(S)} = \frac{5}{36}$

d) The probability of getting that the sum of the dice is a prime number

$P(E) =\frac{n(E)}{n(S)} = \frac{8}{36} = \frac{2}{9}$

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

        $P(E) =\frac{n(E)}{n(S)} = \frac{3}{36} = \frac{1}{12}$

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

$P(E) =\frac{n(E)}{n(S)} = \frac{5}{36}$

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

$P(E) =\frac{n(E)}{n(S)} = \frac{8}{36} = \frac{2}{9}$

Conclusion:-

 a) The size of the sample two dice so n= 2

b) The probability of getting a sum 10 on the two dice $P(E) = \frac{3}{36} = \frac{1}{12}$

c)  The probability of getting a sum of the dice at least 10 on the two dice  

$P(E) =\frac{n(E)}{n(S)} = \frac{5}{36}$

d) The probability of getting that the sum of the dice is a prime number

$P(E) =\frac{n(E)}{n(S)} = \frac{8}{36} = \frac{2}{9}$