Question

The odds against a certain event are 5 to 3 and the odds in favour of another event, independent of the former are 7 to 5. Find the chance that at least one of the event will happen. Chp-probability , plz , plz answer

Answer 1

Add all the odd numbers first and then substitute the value.

Answer 2

Option(C) is correct

Let probability of the first event taking place be A and probability of the second event taking place be B.

Then,

P(A)P(B)=35+3=3/8=77+5=7/12

The required event can be defined as that A takes place and B does not take place (A or B takes place and A does not take place or A takes place and B takes place.)

=[P(A){1−P(B)}]+[P(B){1−P(A)}]+[P(A)P(B)]

=[38×512]+[58×712]+[38×712]

=15+35+2196

=7196