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Home  >>  CBSE XI  >>  Math  >>  Probability

A die is thrown,find the probability of following event : A number greater than or equal to 3 will appear

$\begin{array}{1 1}(A)\;\large\frac{2}{3}\\(B)\;\large\frac{1}{2}\\(C)\;\large\frac{1}{5}\\(D)\;\text{None of these}\end{array} $

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1 Answer

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  • The formula used to calculate is Probability $P=\large\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}=\frac{n(E)}{n(S)}$
Step 1:
Given a die is thrown
$\therefore$ The sample space S={1,2,3,4,5,6}
Let E be the event greater than equal to 3 will appear.
$\therefore$ E={3,4,5,6}
Step 2:
Probability =$\large\frac{n(E)}{n(S)}$
$\Rightarrow \large\frac{4}{6}$
$\Rightarrow \large\frac{2}{3}$
Hence (A) is the correct answer.
answered Jun 30, 2014 by sreemathi.v
 

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