Question

A dice is thrown 80 times. If the probability of having an even number is 7/10 then how many times an odd number appears on dice?​

Answer 1

There are only 2 probabilities here one is of odd and another is of even ....

let p(even) = 7/10 so p(odd) =X

p(even) + p(odd) = 1

7/10+X = 1

x = 1-7/10

x= 3/10

p(odd) = 3/10

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if u want to be objective

by logic u can see that out of 10 , 7 are even so 3 out of 10 will be odd .....

hope it helps .....ask if any doubt

Answer 2

the previous answer is wrong

so,

let s be the sample space

n(s) =80

let A be the event of getting even number

p(A) =n(A) /n(s)

7/10= n(A) /80

n(A) = 7/10*80

= 7*8

=56

let B be the event of getting odd number

n(B) = n(s) - n(A)

= 80 - 56

= 24

so, 24 times an odd number appears on dice