Question
For a biased dice, the probability for the different faces to turn up are
| Face |
1 |
2 |
3 |
4 |
5 |
6 |
| P |
0.10 |
0.32 |
0.21 |
0.15 |
0.05 |
0.17 |
The dice is tossed and it is told that either the face $$1$$ or face $$2$$ has shown up, then the probability that it is face $$1$$, is :
A.
$$\frac{{16}}{{21}}$$
B.
$$\frac{1}{{10}}$$
C.
$$\frac{5}{{16}}$$
D.
$$\frac{5}{{21}}$$
Answer :
$$\frac{5}{{21}}$$
Solution :
Let $$E :$$ ‘face $$1$$ comes up’ and
$$F :$$ ‘face $$1$$ or $$2$$ comes up’
$$\eqalign{
& \Rightarrow E \cap F = E\,\,\,\,\,\,\left( {\because \,E \subset F} \right) \cr
& \therefore \,P\left( E \right) = 0.10{\text{ and}} \cr
& P\left( F \right) = P\left( 1 \right) + P\left( 2 \right) \cr
& = 0.10 + 0.32 \cr
& = 0.42 \cr} $$
Hence, required probability
$$\eqalign{
& = P\left( {\frac{E}{F}} \right) \cr
& = \frac{{P\left( {E \cap F} \right)}}{{P\left( F \right)}} \cr
& = \frac{{P\left( E \right)}}{{P\left( F \right)}} \cr
& = \frac{{0.10}}{{0.42}} \cr
& = \frac{5}{{21}} \cr} $$