Question

The odds in favor of an event a are 3:4. the odds against another independent event b are 7:4. what is the probability that at least one of the events will happen

Answer 1

answer is 4.....
hence there are 4 favourable outcomes possible as 7/4 – 3/7 the answer is 4/7 so there are probabilityof getting 4 results of the event.
hope the answer is correct......
hopefully I may have helped you......

Answer 2

Answer:

The probability that at least one of the events will happen is $\frac{7}{11}$

Step-by-step explanation:

Given : The odds in favor of an event a are 3:4. the odds against another independent event b are 7:4.

To find : What is the probability that at least one of the events will happen?

Solution :

The odds in favor of an event a are 3:4.

i.e. $P(A)=\frac{3}{7}$

The odds against another independent event b are 7:4.

i.e. $P(B)=\frac{4}{11}$

Now we write there compliments,

$P(A')=\frac{4}{7}$

$P(B')=\frac{7}{11}$

The probability that at least one of the events will happen is given by,

$\text{P(at least one of A and B)} = 1 -P(A')P(B')$

$\text{P(at least one of A and B)} = 1 -(\frac{4}{7})(\frac{7}{11})$

$\text{P(at least one of A and B)} = 1 -\frac{4}{11}$

$\text{P(at least one of A and B)} = \frac{11-4}{11}$

$\text{P(at least one of A and B)} = \frac{7}{11}$

Therefore, The probability that at least one of the events will happen is $\frac{7}{11}$