A die is thrown once. Find the probability of getting:a)an odd numberb)a number greater than 4c)a number between 2 and 6.plz solve it....
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Answered by
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a) 3/6=1/2
b)2/6=1/3
c)3/6=1/3
b)2/6=1/3
c)3/6=1/3
Answered by
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n(S) = 6
A = { an odd number }
A = { 1,3,5 }
n(A) = 3
p(A) = n(A)/n(S) = 3/6 (or) 1/2
B = { number greater than 4 }
B = { 5,6 }
n(B) = 2
p(B) = n(B)/n(S) = 2/6 (or) 1/3
C = { number between 2 and 6 }
C = { 3,4,5 }
n(C) = 3
p(C) = n(C)/n(S) = 3/6 (or) 1/2
#Hassan
A = { an odd number }
A = { 1,3,5 }
n(A) = 3
p(A) = n(A)/n(S) = 3/6 (or) 1/2
B = { number greater than 4 }
B = { 5,6 }
n(B) = 2
p(B) = n(B)/n(S) = 2/6 (or) 1/3
C = { number between 2 and 6 }
C = { 3,4,5 }
n(C) = 3
p(C) = n(C)/n(S) = 3/6 (or) 1/2
#Hassan
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