Question

Of all the subsets of {$1,2,3,4,5,6,7$} a non-void subset is randomly selected. The probability that it does not contain two consecutive numbers is equal to

• A. $\frac{33}{128}$
• B. $\frac{35}{128}$
• C. $\frac{68}{128}$
• D. $noneofthese$

Answer 1

The correct option is D $noneofthese$

We need to choose the subsets from $1,2,3,4,5,6,7$
Total number of non void subsets $=2^7-1=64\times 2-1=127$
When there is only one element in set, number of ways of selecting a set ${=}^7{C}_1=7$
Favourable Cases : $\left\{1\right\},\left\{2\right\},\left\{3\right\},\left\{4\right\},\left\{5\right\},\left\{6\right\},\left\{7\right\}$
When there are only 2 elements in a set, number of favourable sets ${=}^7{C}_2-6=15$
Favourable Cases : $\{1,3\},\{1,4\},\{1,5\},\{1,6\},\{1,7\},\{2,4\},\{2,5\},\{2,6\},\{2,7\},\{3,5\},\{3,6\},\{3,7\},\{4,6\},\{4,7\},\{5,7\}$
When there are 3 elements in a set, number of favourable cases ${=}^7{C}_3-(5+4\times 2+3\times 4)=10$
Favourable Cases: $\{1,3,5\},\{1,3,6\},\{1,3,7\},\{1,4,6\},\{1,4,7\},\{1,5,7\},\{2,4,6\},\{2,4,7\},\{2,5,7\},\{3,5,7\}$
When there are only 4 elements in the set, number of favourable cases $=1$
Favourable case : $\{1,3,5,7\}$
It is not possible to select subsets for more than four elements without consecutive numbers.
Total number of favourable ways ${=}^7{C}_1+{(}^7{C}_2-6){+}^7{C}_1-(5+(4\times 2+3\times 4))+1$$=33$
Required probability $=\frac{33}{127}$