Question

A pair of dice is rolled. Find the probability of getting the sum as atmost 7?

Answer 1

Answer:

Step-by-step explanation:

Answer 2

The probability of getting sum of number on its face as atmost 7=$\frac{7}{12}$

Step-by-step explanation:

When two dice are throws

Sample space=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)}

Sum of number on pair of dice as atmost 7

(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(3,1),(3,2),(3,3),(3,4),(4,1),(4,2),(5,1),(4,3),(5,2),(6,1)=21

Total result in one dice=6

Total cases in two dice =$6^2=36$

Favorable cases=21

The probability of getting sum of number on its face as atmost 7=$\frac{21}{36}$

The probability of getting sum of number on its face  as atmost 7=$\frac{7}{12}$

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