Which of the following statement is/are correct?

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Three coins are tossed once. At least two of them must land the same way. No matter whether they land heads or tails, the third coin is equally likely to land either the same way or oppositely. So, the chance that all the three coins land the same way is $\frac{1}{2}$ .

Let $0 < P (B) < 1$ and $P (\frac{A}{B}) = P (\frac{A}{\bar{B}}) .$ Then A and B are dependent.

Suppose an urn contains 'w' white and 'b' black balls and a ball is drawn from it and is replaced along with 'd' additional balls of the same colour. Now a second ball is drawn from it. The probability that the second drawn ball is white is independent of the value of 'd'.

A, B, C simultaneously satisfy

$\begin{array}{l} P (A \cap B \cap C) = P (A) P (B) P (C) \\ P (A \cap B \cap \bar{C}) = P (A) P (B) P (\bar{C}) \\ P (A \cap \bar{B} \cap \bar{C}) = P (A) P (\bar{B}) P (\bar{C}) \\ P (\bar{A} \cap B \cap C) = P (\bar{A}) P (B) P (C) \end{array}$

Then A, B, C are independent.

answer is C, D.

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Detailed Solution

a) false
$P (T T T o r H H H) = \frac{1}{8} + \frac{1}{8} + \frac{1}{4}$
b) $\frac{P (A \cap B)}{P (B)} = \frac{P (A \cap B)}{1 - P (B)} = \frac{P (A) - P (A \cap B)}{1 - P (B)}$
$P (A \cap B) [1 - P (B)] = P (B) P (A) - P (B) P (A \cap B) P (A \cap B) = P (A) P (B)$
Hence the given statement is false
c) Let $E_{1}$ be the white ball is drawn in first drawn $E_{2}$ be the ball is drawn in second drawn E be the event that white ball is drawn in second draw.
$∴ P (E) = P (E_{1}) P (\frac{E}{E_{1}}) + P (E_{2}) P (\frac{E}{E_{2}}) = (\frac{d + w}{w + b + d}) (\frac{w}{w + b}) + (\frac{w}{w + b + d}) + (\frac{b}{w + b}) = (\frac{w}{w + b}) [\frac{d + w}{w + b + d} + \frac{b}{w + b + d}]$
$= \frac{w}{w + b}$ which is independent of ‘d’
d) To prove that A, B, C are pair wise independent only.
Now $P (A \cap B) = P ((A \cap B \cap \bar{C}) \cup (A \cap B \cap C)$
$= P (A \cap B \cap \bar{C}) = P (A \cap B \cap C)$
$= P (A) P (B) P (\bar{C}) + P (A) P (B) P (C)$ GIVEN
$= P (A) P (B) [P (\bar{C}) + P (C)] = P (A) P (B)$
Similarly, for the other two hence this statement is correct