let Roll a pair of fair dice. Let E be the outcome that the sum is odd, the sum

Question

let Roll a pair of fair dice. Let E be the outcome that the sum is odd, the sum is 7, let G be the event that at least one die is a 4, let H be the event that both dice are the same, and I be the event that the first die is a 3. a. Are events E and mutually exclusive? Please justify your reasoning. b. Are events E and H mutually exclusive? Please justify your reasoning c. Find P(EIF) e. Are even f. Are events E and F independent? Please justify your reasoning with a calculation. F be the event that 2. d. Find P(FIE) nts E and independent? Please justify your reasoning with a calculation

Explanation / Answer

a)

as there are events {(4,1),(4,3)...} where one die shows a 4 (event G) and sum on dice are odd (event E)

therefore E and G contain common events ; theefore E and G are not mutually exclusive

b)

as event where both dies are same (event H ) ; let first die shows a number x and second die shows number x ; therfore sum 2x will always be even.

hencce event in H can not be common with event in E ; therefore E and H are mutually exclusive

c)

total number of evnets on two dice =6*6 =36

total ways sum is 7 ={(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)} total 6 events

therefore P(F) =P(sum is 7) =6/36 =1/6

P(E n F) ={(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)} (as sum is odd and  7) =1/6 =P(F)

hence P(E|F )=P(E n F)/P(F) =(1/6)/(1/6) =1

P(E) =1/2 (As sum on dice is odd and even in equal number of cases)

P(F|E) =P(E n F)/P(E) =(1/6)/(1/2)=1/3

e)

here P(I) =6/36=1/6 (as there are 6 events where first die shows 3 and second shows from 1 to 6)

P(E n I) =3/36=1/12 {(3,2),(3,4),(3,6) } 3 events where sum is odd and first die shows a 3)

as P(E)*P(I) =(1/2)*(1/6) =1/12 =P(E n I) ; therefore E and I are independent

f)

P(E)*P(F) =(1/2)*(1/6)=1/12 which is not equal to P(E n F); theefore E and F are not independent

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