Question

In a single throw of two dice find the probability that neither a doublet nor a total of 1 will apoear

Answer 1

Answer:

Probability that neither a doublet nor a total of 1 will appear is 5/6

Step-by-step explanation:

let the event in which neither a doublet nor a total of 1 will appear be E

and sample space be S

S=

(1,1),(1,2),(1,3),(1,4),(1.5),(1,6)

(2,1),(2,2),(2,3),(2,4),(2.5),(2,6)

(3,1),(3,2),(3,3),(3,4),(3.5),(3,6)

(4,1),(4,2),(4,3),(4,4),(4.5),(4,6)

(5,1),(5,2),(5,3),(5,4),(5.5),(5,6)

(6,1),(6,2),(6,3),(6,4),(6.5),(6,6)

n(S)=36

E=

(1,2),(1,3),(1,4),(1.5),(1,6)

(2,1),(2,3),(2,4),(2.5),(2,6)

(3,1),(3,2),(3,4),(3.5),(3,6)

(4,1),(4,2),(4,3),(4.5),(4,6)

(5,1),(5,2),(5,3),(5,4),(5,6)

(6,1),(6,2),(6,3),(6,4),(6.5)

n(E)=30

P(E)=n(E)/n(S)=30/36=5/6

(hey bro i am extremely sorry by mistakely i have rated my own answer please forgive me it is by mistake)