Let A and B be two events such that P(A) = 0.6, P(B) = 0.2, P(A|B) = 0.5. Then text{P}(overline{text{A}}|overline{text{B}}) equals.

Q: Let A and B be two events such that P(A) = 0.6, P(B) = 0.2, P(A|B) = 0.5. Then $\text{P}(\overline{\text{A}}|\overline{\text{B}})$ equals.

Given that, $\text{P(A)}=0.6,\text{P(B)}=0.2,\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=0.5$ Consider, $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=0.5$ $\Rightarrow\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}=0.5$ $\Rightarrow\frac{\text{P}(\text{A}\cap\text{B})}{0.2}=0.5$ $\Rightarrow \text{P}(\text{A}\cap\text{B})=0.1$ $\Rightarrow \text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})=0.1$ $\Rightarrow \text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-0.1$ $\Rightarrow\text{P}(\text{A}\cup\text{B})=0.7$ Now, $\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=\text{P}(\overline{\text{A}\cup\text{B}})$ $\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=1-\text{P}(\text{A}\cup\text{B})$ $\Rightarrow \text{P}(\overline{\text{A}}\cap\overline{\text{B}})=0.3$ To find $\text{P}\Big(\frac{\overline{\text{A}}}{\overline{\text{B}}}\Big)=\frac{\text{P}(\overline{\text{A}\cap\text{B}})}{\text{P}(\overline{\text{B}})}$ $\text{P}\Big(\frac{\overline{\text{A}}}{\overline{\text{B}}}\Big)=\frac{0.3}{0.8}$ $\text{P}\Big(\frac{\overline{\text{A}}}{\overline{\text{B}}}\Big)=\frac{3}{8}$

M.C.Q (1 Marks)

GSEB - English Medium

Let A and B be two events such that P(A) = 0.6, P(B) = 0.2, P(A|B) = 0.5. Then $\text{P}(\overline{\text{A}}|\overline{\text{B}})$ equals.

A$\frac{1}{10}$
B$\frac{3}{10}$
C$\frac{3}{8}$
D$\frac{6}{7}$

STD 12 Science
Maths
Probability
Exemplar

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