Math, asked by sannyashi7446, 12 days ago

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 KBurrell02
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 kashyapsingh52
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

Similar questions