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# A die is thrown,find the probability of following event : A number more than 6 will appear

$\begin{array}{1 1}(A)\;\large\frac{1}{2}\\(B)\;\large\frac{3}{5}\\(C)\;0\\(D)\;\phi\end{array}$

Toolbox:
• 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}
A number more than 6 cannot appear on a die because there is no number greater than 6 on a die.
$\therefore$ Let E be the event a number more than 6 will appear.
$\therefore E=\{ \}=\phi$
Step 2:
Probability $=\large\frac{n(E)}{n(S)}=\frac{0}{6}$$=0$
Hence (C) is the correct answer.