Q74 / 180 · COMEDK 2024 (Afternoon Shift) · Mathematics
A and B are two independent events. The probability of their simultaneous occurrence is $\dfrac{1}{8}$ and the probability that neither of them occurs is $\dfrac{3}{8}$. Then their individual probabilities are
- A$\dfrac{3}{8} \text { and } \dfrac{1}{8}$
- B$\dfrac{5}{8} \text { and } \dfrac{1}{4}$
- C$\dfrac{3}{4} \text { and } \dfrac{1}{2}$
- Correct answerD$\dfrac{1}{2} \text { and } \dfrac{1}{4}$
Solution
Let $P(A) = x$ and $P(B) = y$. Since $A$ and $B$ are independent events, $P(A \cap B) = P(A)P(B) = xy = \dfrac{1}{8}$.
The probability that neither occurs is $P(A^{c} \cap B^{c}) = P(A^{c})P(B^{c}) = (1-x)(1-y) = \dfrac{3}{8}$.
Expanding the second equation: $1 - x - y + xy = \dfrac{3}{8}$.
Substituting $xy = \dfrac{1}{8}$ into the equation: $1 - (x+y) + \dfrac{1}{8} = \dfrac{3}{8}$.
Rearranging gives $x+y = 1 + \dfrac{1}{8} - \dfrac{3}{8} = 1 - \dfrac{2}{8} = 1 - \dfrac{1}{4} = \dfrac{3}{4}$.
We have the sum $x+y = \dfrac{3}{4}$ and the product $xy = \dfrac{1}{8}$. These are the roots of the quadratic equation $t^2 - (x+y)t + xy = 0$, which is $t^2 - \dfrac{3}{4}t + \dfrac{1}{8} = 0$.
Multiplying by 8: $8t^2 - 6t + 1 = 0$.
Factoring the quadratic: $8t^2 - 4t - 2t + 1 = 0 \Rightarrow 4t(2t-1) - 1(2t-1) = 0 \Rightarrow (4t-1)(2t-1) = 0$.
The roots are $t = \dfrac{1}{4}$ and $t = \dfrac{1}{2}$. Thus, the probabilities are $\dfrac{1}{2}$ and $\dfrac{1}{4}$.
Answer: $\dfrac{1}{2} \text { and } \dfrac{1}{4}$
The probability that neither occurs is $P(A^{c} \cap B^{c}) = P(A^{c})P(B^{c}) = (1-x)(1-y) = \dfrac{3}{8}$.
Expanding the second equation: $1 - x - y + xy = \dfrac{3}{8}$.
Substituting $xy = \dfrac{1}{8}$ into the equation: $1 - (x+y) + \dfrac{1}{8} = \dfrac{3}{8}$.
Rearranging gives $x+y = 1 + \dfrac{1}{8} - \dfrac{3}{8} = 1 - \dfrac{2}{8} = 1 - \dfrac{1}{4} = \dfrac{3}{4}$.
We have the sum $x+y = \dfrac{3}{4}$ and the product $xy = \dfrac{1}{8}$. These are the roots of the quadratic equation $t^2 - (x+y)t + xy = 0$, which is $t^2 - \dfrac{3}{4}t + \dfrac{1}{8} = 0$.
Multiplying by 8: $8t^2 - 6t + 1 = 0$.
Factoring the quadratic: $8t^2 - 4t - 2t + 1 = 0 \Rightarrow 4t(2t-1) - 1(2t-1) = 0 \Rightarrow (4t-1)(2t-1) = 0$.
The roots are $t = \dfrac{1}{4}$ and $t = \dfrac{1}{2}$. Thus, the probabilities are $\dfrac{1}{2}$ and $\dfrac{1}{4}$.
Answer: $\dfrac{1}{2} \text { and } \dfrac{1}{4}$
Hence, the correct answer is option: 'D'.