An unbiased die is thrown twice. Let the event $A$ be 'odd number on the first throw' and $B$ the event 'odd number on the second throw'. Check the independence of the events $A$ and $B$.
(A) If all the $36$ elementary events of the experiment are considered to be equally likely,we have
$P(A) = \frac{18}{36} = \frac{1}{2}$ and $P(B) = \frac{18}{36} = \frac{1}{2}$
Also,$P(A \cap B) = P(\text{odd number on both throws})$
$= \frac{9}{36} = \frac{1}{4}$
Now,$P(A) \times P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
Clearly,$P(A \cap B) = P(A) \times P(B)$
Thus,$A$ and $B$ are independent events.
$P(A) = \frac{18}{36} = \frac{1}{2}$ and $P(B) = \frac{18}{36} = \frac{1}{2}$
Also,$P(A \cap B) = P(\text{odd number on both throws})$
$= \frac{9}{36} = \frac{1}{4}$
Now,$P(A) \times P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
Clearly,$P(A \cap B) = P(A) \times P(B)$
Thus,$A$ and $B$ are independent events.
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View SolutionConsider two events $A$ and $B$ such that $P(A) = \frac{1}{4}$,$P(B/A) = \frac{1}{2}$,$P(A/B) = \frac{1}{4}$. For each of the following statements,which is true?
$I.$ $P(A^c/B^c) = \frac{3}{4}$
$II.$ The events $A$ and $B$ are mutually exclusive
$III.$ $P(A/B) + P(A/B^c) = 1$
$I.$ $P(A^c/B^c) = \frac{3}{4}$
$II.$ The events $A$ and $B$ are mutually exclusive
$III.$ $P(A/B) + P(A/B^c) = 1$
Difficult
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