Correct Answer - Option 4 :
\(\frac17\)
Concept:
\(\rm P(A')=1-P(A)\)
\(\rm P(A\cap B')=P(A)-P(A\cap B)\)
\(\rm P(\frac AB)=\frac{P(A \; \cap \; B)}{P(B)}\)
\(\rm P(A\cup B)=P(A)+P(B)-P(A\cap B)\)
Calculation:
Here, P(A') = 0.4, P(B) = 0.4 and P(A ∩ B') = 0.5
\(\rm P(A)=1-P(A')\\=1-0.4\\=0.6\)
\(\rm P(B')=1-P(B)\\=1-0.4\\=0.6\)
Now, \(\rm P(\frac{B}{A\;\cup\; B'})=\frac{P(B\;\cap\;(A\;\cup\; B'))}{P(A\;\cup \;B')}\) (∵\(\rm P(\frac AB)=\frac{P(A\cap B)}{P(B)}\))
\(\rm =\frac{P(B\;\cap \;A)}{P(A\;\cup \; B')}\) (∵ \(\rm (A\cup B')= A\))
As we know, \(\rm P(A\cap B')=P(A)-P(A\cap B)\) and \(\rm P(A\cup B)=P(A)+P(B)-P(A\cap B)\)
\(\rm =\frac{P(A)-P(A\;\cap\; B')}{P(A)+P(B')-P(A\;\cap\; B')}\)
\(=\rm \frac{0.6-0.5}{0.6+0.6-0.5}\\=\frac{0.1}{0.7}\)
\(=\frac17\)
Hence, option (4) is correct.
