Question

Let A and B be two independent events such that the odds in favour of A and B are $1:1$ and $3:2$ respectively .Then the probability that only one of the two occurs is

• A. $0.6$
• B. $0.7$
• C. $0.8$
• D. $0.5$
• E. $0.4$

Answer 1

Answer: (D)

$0.5$

Answer 2

Given that:

Fir the given two events A and B

Odds in favor of A = 1:1.

Odds in favor of B = 3:2.

Therefore, probability of A P(A) =$\dfrac{1}{(1 + 1)} = \dfrac{1}{2}$ and probability of B P(B) = $\dfrac{3}{(3 + 2)} = \dfrac{3}{5}$.

As per the question we need to find the probability that only one of the two events occurs. This means either event A occurs (but not B), or event B occurs (but not A).

Probability of only A occurring = P(A) × (1 - P(B)) =$(\dfrac{1}{2}) × (1 - \dfrac{3}{5})$

$= (\dfrac{1}{2}) × (\dfrac{2}{5})$

$= \dfrac{1}{5}$

Probability of only B occurring = P(B) × (1 - P(A)) = $(\dfrac{3}{5}) × (1 - \dfrac{1}{2})$

$= (\dfrac{3}{5}) × (\dfrac{1}{2})$

$= \dfrac{3}{10}$

Now, to find the total probability that only one of the two events occurs, we can write,

Total probability = Probability of only A + Probability of only B

$= \dfrac{1}{5} + \dfrac{3}{10}$

$= \dfrac{1}{2}$

So, the probability that only one of the two events A or B occurs is $\dfrac{1}{2}$.

Hence the correct answer is $0.5$ (_Ans.)