Question

The odds against a certain event are 5 : 2 and odds in favour of another independent event are 6 : 5. Find the chance that at least one of the events will happen.

Answer 1

Answer:

47/77

Step-by-step explanation:

Hi,

Odds against a certain event =

No of unfavorable outcome/Total number of outcome

Let E be any event whose odds against the event are a : b

Number of Unfavorable Outcomes/Number of Favorable

Outcomes = a/b

= 1 + Number of Unfavorable Outcomes/Number of Favorable

Outcomes

= 1 + a/b

(Number of Favorable Outcomes  + Number of Unfavorable

Outcomes)/Number of Favorable Outcomes = (a + b)/b

Total number of outcomes/Number of favorable outcomes

= (a + b)/b

P(E) = Number of Favorable Outcomes/Total number of

outcomes

= b/(a + b)

Let 'A' be the event whose add against it are 5 : 2

So, P(A) = 2/(2 + 5) = 2/7

Let 'B' be the event whose add against it are 6 : 5

So, P(B) = 5/(6 + 5) = 5/11

Given that A and B are independent, so

P(A ∩ B) = P(A)P(B) = 2/7*5/11

 = 10/77

Chance that at least one of the events A or B will happen is

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

= 2/7 + 5/11 - 10/77

= 47/77

Hope, it helped !

Answer 2

Answer:52/77

Step-by-step explanation:

Let

A:first event will happen

B:second event will happen

Given-P(not A)=5/7

P(B)=6/11

Then,P(A)=1-5/7=2/7

To find: P(A union B)=?

P(A intersection B)=P(A)*P(B)= 2/7*6/11=12/77

P(A union B)=P(A)+P(B)-PA intersection B)

=2/7+6/11-12/77

=52/77