PUMPA - SMART LEARNING
எங்கள் ஆசிரியர்களுடன் 1-ஆன்-1 ஆலோசனை நேரத்தைப் பெறுங்கள். டாப்பர் ஆவதற்கு நாங்கள் பயிற்சி அளிப்போம்
Book Free DemoTheorem:
If \(A\) and \(B\) are two events associated with a random experiment, then prove that:
(i) \(P(A \cap \overline B)\) \(=\) \(P(\text{only A})\) \(=\) \(P(A) - P(A \cap B)\)
(ii) \(P(\overline A \cap B)\) \(=\) \(P(\text{only B})\) \(=\) \(P(B) - P(A \cap B)\)
Proof for the theorem:
Statement (i):
The Venn diagram representing \(A \cap \overline B\) is as follows:
To prove: \(P(A \cap \overline B)\) \(=\) \(P(A) - P(A \cap B)\)
Proof for statement (i):
By the distributive property of sets, we have:
1. \((A \cap B) \cup (A \cap \overline B)\) \(=\) \(A \cap (B \cup \overline B)\)
Here, \(B \cup \overline B = S\).
Thus, \((A \cap B) \cup (A \cap \overline B)\) \(=\) \(A \cap S\)
Hence, \((A \cap B) \cup (A \cap \overline B)\) \(=\) \(A\).
2. \((A \cap B) \cap (A \cap \overline B)\) \(=\) \(A \cap (B \cap \overline B)\)
Here, \(B \cap \overline B = \phi\).
Thus, \((A \cap B) \cap (A \cap \overline B)\) \(=\) \(A \cap \phi\)
Hence, \((A \cap B) \cap (A \cap \overline B)\) \(=\) \(\phi\).
Therefore, from the above properties it is clear that the events \((A \cap B)\) and \((A \cap \overline B)\) are mutually exclusive and its union is set \(A\).
So, \(P(A)\) \(=\) \(P\left[(A \cap B) \cup (A \cap \overline B)\right]\)
\(\Rightarrow\) \(P(A)\) \(=\) \(P(A \cap B) + P(A \cap \overline B)\)
\(\Rightarrow\) \(P(A \cap \overline B)\) \(=\) \(P(A)\) \(-\) \(P(A \cap B)\)
Therefore, \(P(A \cap \overline B)\) \(=\) \(P(\text{only A})\) \(=\) \(P(A) - P(A \cap B)\).
Hence, proved.
Statement (ii):
The Venn diagram representing \(\overline A \cap B\) is as follows:
To prove: \(P(\overline A \cap B)\) \(=\) \(P(B) - P(A \cap B)\)
Proof for statement(ii):
By the distributive property of sets, we have:
1. \((A \cap B) \cup (\overline A \cap B)\) \(=\) \((A \cup \overline A) \cap B\)
Here, \(A \cup \overline A = S\).
Thus, \((A \cap B) \cup (\overline A \cap B)\) \(=\) \(S \cap B\)
Hence, \((A \cap B) \cup (\overline A \cap B)\) \(=\) \(B\).
2. \((A \cap B) \cap (\overline A \cap B)\) \(=\) \((A \cap \overline A) \cap B\)
Here, \(A \cap \overline A = \phi\).
Thus, \((A \cap B) \cap (\overline A \cap B)\) \(=\) \(\phi \cap B\)
Hence, \((A \cap B) \cap (\overline A \cap B)\) \(=\) \(\phi\).
Therefore, from the above properties it is clear that the events \((A \cap B)\) and \((\overline A \cap B)\) are mutually exclusive and its union is set \(B\).
So, \(P(B)\) \(=\) \(P\left[(A \cap B) \cup (\overline A \cap B)\right]\)
\(\Rightarrow\) \(P(B)\) \(=\) \(P(A \cap B) + P(\overline A \cap B)\)
\(\Rightarrow\) \(P(\overline A \cap B)\) \(=\) \(P(B)\) \(-\) \(P(A \cap B)\)
Therefore, \(P(\overline A \cap B)\) \(=\) \(P(\text{only B})\) \(=\) \(P(B) - P(A \cap B)\).
Hence, proved.
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