If A, B and C are three events, then which of the following is incorrect?

P ( A and at least one of B C, occurs) ≥ $P(A∩B) +P(A ∩ C)$

We have,

P (Exactly two of A, B, C occur)

$= P(A ∩ B ∩ \overline{C})+ P(A ∩ \overline{B} ∩ C) + P( \overline{A} ∩ B ∩ C)$

$= P( A ∩ B) - P( A ∩ B ∩ C)+ P( A ∩ C) - P( A ∩ B ∩ C)$ + P(B ∩ C - P ( A ∩ B ∩ C)$

$= P(A ∩ B)+ P( B ∩  C) + P( A ∩ C) - 3P (A ∩ B ∩ C) ≤ P(A ∩ B) + P( B ∩ C) + P( A ∩ C)$

Also,

$P(A ∪ B ∪ C)$

$= P (A ∪ B) + P(C) - P  \begin{Bmatrix}(A ∪ B) ∩ C\end{Bmatrix}$

$≤ P(A ∪ B) + P(C).$

$≤  P(A) +P(B) +P(C)$              $[∵P(A ∪ B) ≤  P(A) + P(B) ]$

Now,

P (Exactly one of A, B, C occurs)

$P(A ∩ \overline{B} ∩ \overline{C})+ P(\overline{A} ∩ \overline{B} ∩ C)+P(\overline{A} ∩ B ∩ \overline{C})$

$ = P(A ∩ \overline {B ∪ C}) + P(\overline {A ∪ B}∩ C) + P( B ∩\overline {A ∪ C})$

$= P(A) - P \begin{Bmatrix}A ∩(B ∪ C) \end{Bmatrix} +P(C)-P\begin{Bmatrix} C ∩ ( A ∪B)\end{Bmatrix} + P(B) - P\begin{Bmatrix} B ∩(A ∪ C)\end{Bmatrix}$

$= P(A) - P \begin{Bmatrix}(A ∩B )∪ (A ∩ C) \end{Bmatrix} +P(C)-P\begin{Bmatrix} (C ∩ A)∪(C∩B)\end{Bmatrix} + P(B) - P\begin{Bmatrix} (B ∩A) ∪ (B  ∩ C)\end{Bmatrix}$

$= P(A) +P(B) +P(C) -2P (A ∩ B) -2P( B ∩ C) - 2P ( A ∩ C) + 3P ( A ∩ B ∩ C)$

$=[P(A) +P(B) +P(C) -P(A ∩ B) -P(B ∩ C) -P (A ∩ C)]-[P(A ∩ B)+P(B ∩ C) + P(A ∩ C) -3P (A ∩ B ∩ C)]$

$= P(A) +P(B) +P(C) -P(A ∩ B) - P(B ∩ C)-P (A ∩ C) $ -P (Exactly two of A, B , C occur)

$≤ P(A) +P(B) +P(C) -P(A ∩ B) -P(B ∩ C) -P( A ∩ C)$

Finally,

P(A and atleast one of B, C occurs)

$=P[A ∩ (B ∪ C)]$

$= P[(A ∩ B) ∪ (A ∩ C)]$

$= P[(A ∩ B) +P (A ∩ C)-P [(A ∩ B) ∩ (A ∩ C)]$

$= P(A ∩ B) + P(A ∩ C) - P(A ∩ B ∩ C)$

$≤ P(A ∩ B) + P(A ∩ C)$

So, option (d) is incorrect.