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.
Answers
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: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