Question

Which of the following cannot be valid assignment of probability for elementary events or outcomes of sample space

$S=\{w_1,w_2,w_3,w_4,w_5,w_6,w_7\}$:
Elementary events :

(i) $\begin{array}{ccccccc}{w}_1& w_2& w_3& w_4& w_5& w_6& w_7\\ 0.1& 0.01& 0.05& 0.03& 0.01& 0.2& 0.6\end{array}$

(ii) $\begin{array}{ccccccc}\frac{1}{7}& \frac{1}{7}& \frac{1}{7}& \frac{1}{7}& \frac{1}{7}& \frac{1}{7}& \frac{1}{7}\end{array}$

(iii) $\begin{array}{ccccccc}0.7& 0.6& 0.5& 0.4& 0.3& 0.2& 0.1\end{array}$

(iv) $\begin{array}{ccccccc}\frac{1}{14}& \frac{1}{14}& \frac{1}{14}& \frac{1}{14}& \frac{1}{14}& \frac{1}{14}& \frac{1}{14}\end{array}$

Answer 1

The sample space of the given experiment is given by S = {1,2,3,4, 5, 6}
$\therefore$ n(S)=6

(i) Let A be the event of occurrence of a prime number.
Then A = {2, 3, 5}
i.e., n(A) = 3

$\therefore P\left(A\right)=\frac{\text{Number of outcomes favourable to A}}{\text{Total number of possible outcomes}}=\frac{n\left(A\right)}{n\left(S\right)}=\frac{3}{-6}=\frac{1}{2}$

(ii) Let B be the event of occurrence of the number 2 or 4.
Then B = {2, 4}
i.e., n(B) = 2

$\therefore P\left(B\right)=\frac{\text{Number of outcomes favourable to B}}{\text{Total number of possible outcomes}}\frac{n\left(B\right)}{n\left(S\right)}=\frac{2}{6}=\frac{1}{3}$

Let C be the event of occurrence of a multiple of 2 or 3. T
Then C = {2, 3, 4,6}
i.e., n(C)= 4

$\therefore P\left(C\right)=\frac{\text{Number of outcomes favourable to C}}{\text{Total number of possible outcomes}}=\frac{n\left(C\right)}{n\left(S\right)}=\frac{4}{6}=\frac{2}{3}$