The probability of the simultaneous occurrence of two events A and B is p. If the probability that exactly one of A, B occurs is q, then :

Options

A. `P(overline(A))+P(overline(B))=2+2q-p`
B. `P(overline(A))+P(overline(B))=2-2p-q`
C. `P(A cap B // A cup B)=(p)/(p+q)`
D. `P(overline(A) cap overline(B))=1-p-q`

Answer

A

Explanation

We have,
`P(A cap B)= p " and " P(A)+P(B)+P(B)-2P(A cap B)=q`
`implies P(A)+P(B)-2p+q`
`implies P(A)+P(B)=2p+q`
`implies 1-P(overline(A))+1-P(overline(B))=2p+q`
`implies P(overline(A))+P(overline(B))=2-2p-q`
So, alternative (b) is correct.
Now,
`{(A cap B)//(A cup B)}=(P[(A cap B) cap (A cup B)])/(P(A cup B))`
`implies P{(A cap B)//(A cup B)}=(P(A cap B))/(P(A cup B))`
`implies P{(A cap B)//(A cup B)}=(P(A cap B))/(P(A)+P(B)-P(A cap B))`
`implies P{(A cap B)//(A cup B)}=(p)/(2p+q-p)=(p)/(p+q)`
So, alternative (c ) is correct.
Finally,
`P(overline(A) cap overline(B))=P(overline(A cup B))=1-P(A cup B)`
`implies P(overline(A) cap overline(B))=P(overline(A cup B))=1-[P(A)+P(B)-P(A cap B)]`
`implies P(overline(A) cap overline(B))=P(overline(A cup B))=1-[2p+q-p]=1-p-q`
So, alternative (d) is correct.
Hence, alternative (a) is incorrect.