Question

The probabilities of different faces of a biased dice to appear are as follows
$\begin{array}{ccccccc}Facenumber& 1& 2& 3& 4& 5& 6\\ Probability& 0.1& 0.32& 0.21& 0.15& 0.05& 0.17\end{array}$
The dice is thrown and it is known that either the face number 1 or 2 will appear. Then, the probability of the face number 1 to appear is

• A. $\frac{5}{21}$
• B. $\frac{5}{13}$
• C. $\frac{7}{23}$
• D. $\frac{3}{10}$

Answer 1

The correct option is A $\frac{5}{21}$

The probability of the face number 1 to appear
$=\frac{P\left(1\right)}{P\left(1\right)+P\left(2\right)}=\frac{0.1}{0.1+0.32}=\frac{0.1}{0.42}=\frac{10}{42}=\frac{5}{21}$