Question

there are two die A and B both having six faces .Die A has three faces marked as 1.two die faces marked as 2 and one faces marked as 3 .Die B has one one face marked with 1 two face marked with 2 three face marked with 3 both the die thrown randomly once if E be the event of getting sum of number appearing on top faces equal to x and let P(E) be the probability of event E than p(e) is the maximum when X is equal to ???????​

Answer 1

Given :  Tow die A & B having six faces each. Die A - 1 - 3 times , 2 - 2times & 3 - 1 times  Dice B has   1 - 1 times , 2 - 2 times & 3 - 3 times

To find : Which sum has maximum probability

Step-by-step explanation:

Dice A -  { 1 , 1 , 1  , 2 , 2 , 3}

P(1) = 3/6 = 1/2

P(2) = 2/6 = 1/3

P(3) = 1/6

Dice B - { 1 , 2 , 2  , 3 , 3 , 3 )

P(1) = 1/6

P(2) = 1/3

P(3) = 1/2

Sum = { 2 , 3  , 4 , 5 , 6)

Sum 2 = 1 + 1

P(2)  = (1/2)(1/6)  = 1/12

Sum 3  = 1 + 2  or 2 + 1

P(3) = (1/2)(1/3) + (1/3)(1/6)  = 4/18  = 2/9

sum 4 = 1 + 3 , 2 + 2 , 3 + 1

P(4) = (1/2)(1/2)  + (1/3)(1/3) + (1/6)(1/6)  = (9 + 4 + 1)/36 = 14/36  = 7/18

Sum 5 =  2 + 3  or  3 + 2

P(5)  = (1/3)(1/2) + (1/6)(1/3)  = 4/18  = 2/9

Sum = 6  = 3 +3

P(6)  =  (1/6)(1/2)  = 1/12

7/18  is maximum  

=> X = 4

probability of getting sum 4 is maximum

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