Mark tosses a dice twice. What is the probability that the sum of the values obtained in the throws is 9?


Answer:

 

4
36
 

Step by Step Explanation:
  1. Let us assume that S, E and P are the sample space, event, and probability, respectively, of getting the sum of the values obtained in the throws is 9 by tossing a dice twice.

    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) }

    E = {(3,6),(4,5),(5,4),(6,3)}
  2. P(E) =  
    n(E)
    n(S)
     
    (Since, P(E) is the probability of getting an event E, n(E) is the number of elements in the event E, and n(S) is the number of elements in the sample space S )
    P(E) =  
    4
    36
     
  3. Hence, the probability that the sum of the values obtained in the throws is 9 by tossing the dice twice is  
    4
    36
     

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