Question: There are 3 events, E1, E2, and E3, one of which must and only can happen.The odds are 7 to 1 against E1, 5 to 3 against E2. Find the odds against E3.
Answers
Answered by
0
Answer: 9/35 or 35 to 9
Step-by-step explanation: the peobabilities in this case must add to 1. Therefore you can do 1 - 1/7 - 3/5
which is equal to 9/35
Hope this helps
Answered by
0
Answer:
1
Step-by-step explanation:
odds against an event= unfavourable outcomes÷favourable outcomes
probabilty of an event=favourable outcomes÷total outcomes
P(E1)=1÷8
P(E2)=3÷8
p(E1∪E2∪E3)=P(E1)+P(E2)+P(E3)=1 as events are mutually exclusive and exhaustive
p(E3)=1-1/8-3/8=1/2=favourable/total
odds against E3=1/1=unfavourable/favourable
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