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Home  >>  CBSE XI  >>  Math  >>  Probability
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A die is loaded in such a way that each odd number is twice as likely to occur as each even number.Find P(G),where G is the event that a number greater than 3 occurs on a single roll of the die.

$\begin{array}{1 1}(A)\;\large\frac{4}{9}\\(B)\;\large\frac{3}{7}\\(C)\;\large\frac{5}{7}\\(D)\;\large\frac{5}{9}\end{array} $

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  • Required probability =$\large\frac{n(E)}{n(S)}$
Step 1:
A die is rolled such that each odd number occurs twice and even number occurs once.
$\therefore$ Total sample space =$\{1,1,2,3,3,4,5,5,6\}$
P(Greater than 3) to occur.
Step 2:
$\therefore$ Number of favorable outcomes n(E)=$\{4,5,5,6\}$
$\Rightarrow 4$
Step 3:
$\therefore$ Required probability =$\large\frac{n(E)}{n(S)}$
$\Rightarrow \large\frac{4}{9}$
Hence (A) is the correct answer.
answered Jul 7, 2014 by sreemathi.v

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