Question

If $P(\overline{A}\cap B)+P(A\cap \overline{B})=1-k$, $P(\overline{A}\cap C)+P(A\cap \overline{C})=1-2K$, $P(\overline{B}\cap C)+P(\overline{C}\cap B)=1-K$ and $P(A\cap B\cap C)=K^2$ then find the value of P(at least one of A, B, C) is

• A. $>\frac{1}{2}$
• B. $\left[\frac{1}{8},\frac{1}{2}\right]$
• C. $<\frac{1}{4}$
• D. $\frac{1}{4}$

Answer 1

The correct option is A

$>\frac{1}{2}$

Step-1: Apply the formula of Sets:

Given $P(\overline{A}\cap B)+P(A\cap \overline{B})=1-k$, --------(i)

$P(\overline{A}\cap C)+P(A\cap \overline{C})=1-2K$, --------(ii)

$P(\overline{B}\cap C)+P(\overline{C}\cap B)=1-K$ ----------(iii)

and $P(A\cap B\cap C)=K^2$ ---------(iv)

By the formula of Sets, we have

$P(\overline{A}\cap B)=P\left(B\right)-P(A\cap B)$

Apply the above formula to L.H.S of the equations (i)

$$\begin{gathered} P(\overline{A}\cap B)+P(A\cap \overline{B})=1-k \\ P\left(B\right)-P(A\cap B)+P\left(A\right)-P(A\cap B)=1-k \\ P\left(A\right)+P\left(B\right)-2P(A\cap B)=1-K \end{gathered}$$

From the above equation, we can write

$P(A\cap B)=\frac{P\left(A\right)+P\left(B\right)-1+k}{2}$ ….(v)

Apply the above formula to L.H.S of the equations (iI)

$$\begin{gathered} P(\overline{A}\cap C)+P(A\cap \overline{C})=1-2k \\ P\left(C\right)-P(A\cap C)+P\left(A\right)-P(A\cap C)=1-2k \\ P\left(A\right)+P\left(C\right)-2P(A\cap C)=1-2K \end{gathered}$$

From the above equation, we can write

$P(A\cap C)=\frac{P\left(A\right)+P\left(C\right)-1+2k}{2}$ …(vi)

Apply the above formula to L.H.S of the equations (iii)

$$\begin{gathered} P(\overline{B}\cap C)+P(B\cap \overline{C})=1-k \\ P\left(C\right)-P(B\cap C)+P\left(B\right)-P(B\cap C)=1-k \\ P\left(B\right)+P\left(C\right)-2P(B\cap C)=1-k \end{gathered}$$

From the above equation, we can write

$P(B\cap C)=\frac{P\left(B\right)+P\left(C\right)-1+k}{2}$ ….(vii)

Step 2: Find the value of $P(A\cup B\cup C)$

We also have the formula as

$P(A\cup B\cup C)=P\left(A\right)+P\left(B\right)+P\left(C\right)-P(A\cap B)-P(B\cap C)-P(A\cap C)+P(A\cap B\cap C)$

Now substitute the values of (iv), (v), (vi),(vii) in the above equation

$\begin{array}{rcl}\therefore P(A\cup B\cup C)& =& P\left(A\right)+P\left(B\right)-P\left(C\right)-P(A\cap B)-P(B\cap C)-P(A\cap C)+P(A\cap B\cap C)\\ & =& P\left(A\right)+P\left(B\right)-P\left(C\right)-\left[\frac{P\left(A\right)+P\left(B\right)-1+k}{2}\right]-\left[\frac{P\left(B\right)+P\left(C\right)-1+k}{2}\right]-\left[\frac{P\left(A\right)+P\left(C\right)-1+2k}{2}\right]+k^2\\ & =& \frac{2k^2-4k+3}{2}\end{array}$

Step 3: Apply $P(A\cup B\cup C)=P(\mathrm{at}\mathrm{least}\mathrm{one}\mathrm{of}A,B,C)$

Since,

$P(A\cup B\cup C)=\frac{2k^2-4k+3}{2}$

$P(\mathrm{at}\mathrm{least}\mathrm{one}\mathrm{of}A,B,C)=$ $\frac{2k^2-4k+3}{2}$

But $\frac{2k^2-4k+3}{2}>\frac{1}{2}$

Therefore, $P(\mathrm{at}\mathrm{least}\mathrm{one}\mathrm{of}A,B,C)>$ $\frac{1}{2}$

Hence, option(A) is correct.