JEE MAIN - Mathematics (2019 - 11th January Evening Slot - No. 3)
Let S = {1, 2, . . . . . ., 20}. A subset B of S is said to be "nice", if the sum of the elements of B is 203. Then the probability that a randonly chosen subset of S is "nice" is :
$${5 \over {{2^{20}}}}$$
$${7 \over {{2^{20}}}}$$
$${4 \over {{2^{20}}}}$$
$${6 \over {{2^{20}}}}$$
Explanation
We can solve this problem by counting the number of "nice" subsets in the set S = {1, 2, $\ldots$, 20 }, and then dividing that number by the total number of possible subsets of S.
The sum of all elements in S is :
1 + 2 + $\ldots$ + 20 = $\frac{{20 \times 21}}{2}$ = 210
Since a "nice" subset must sum to 203, the elements not in the subset must sum to 210 - 203 = 7.
Now we need to find the ways to make the sum of 7 using the elements of S. The combinations are :
- 1. 7
- 2. 1 + 6
- 3. 2 + 5
- 4. 3 + 4
- 5. 1 + 2 + 4
- 6. 1 + 3 + 3(This doesn't work since 3 is repeated)
- 7. 2 + 2 + 3(This doesn't work since 2 is repeated)
So, there are 5 "nice" subsets.
Since the set S has 20 elements, there are $2^{20}$ possible subsets (including the empty set and the set itself). The probability of randomly choosing a "nice" subset is therefore :
Comments (0)
