Question

In a simultaneous throw of a pair of dice find the probability of getting : (a) a doublet (b) a doublet of even number (c) a sum greater than 7(d) an even numbers on first die (e) a number other than 5 on any die​

Answer 1

Answer:

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

a)E₁=An event of getting a doublet=(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)

   n(E₁)=6

∴P(getting a doublet)=n(E₁)/n(S) =6/36=1/6

b)E₂= An event of getting a doublet of even nos.=(2,2),(4,4),(6,6)

  n(E₂)=3

∴P(getting a doublet of even nos.) =n(E₂)/n(S)=3/36=1/18  

c)E₃=An event of getting a sum greater than 7= (2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6)

 n(E₃)=15

∴P(getting a sum greater than 7) =n(E₃)/n(S)=15/36=5/12

d)E₄=An event of getting an even no. on the first die= (2,1);(2,2);(2,3);(2,4);(2,5);(2,6);(4,1);(4,2);(4,3);(4,4);(4,5);(4,6);(6,1);(6,2);(6,3);(6,4);(6,5);(6,6)  

 n(E₄)=18

∴P(getting an even no. on the first die) =n(E₄)/n(S)=18/36=1/2

e)E₅=An event of getting a no. other than 5 on any die= (1,1);(1,2);(1,3);(1,4);(1,6);(2,1);(2,2);(2,3);(2,4);(2,6);(3,1);(3,2);(3,3);(3,4);(3,6);(4,1);(4,2);(4,3);(4,4);(4,6);     (6,1);(6,2);(6,3);(6,4);(6,6)

 n(E₅)=11

∴P(getting a no. other than 5 on any die) =n(E₅)/n(S)=11/36    

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