Let S = E₁, E₂, ........., E₈ be a sample space of a random experiment such that P(Eₙ) = n/36 for every n = 1, 2, ........, 8. Then the number of elements in the set { A ⊆ S:P(A) ≥ 4/5 } is ___________.

Solution

Maths sir · 3 steps

Turn the probability condition into a sum of indices, then count the small leftover sets instead.

1. Total check. The probabilities add to (1+2+…+8)/36 = 36/36 = 1. So P(A) = (sum of indices n in A)/36.

   sum(A)/36 ≥ 4/5 ⇒ sum(A) ≥ 28.8 ⇒ sum(A) ≥ 29

2. Use the complement. The indices left out of A add to 36 - sum(A). So sum(A) ≥ 29 exactly when the left-out indices add to at most 7. Each such left-out set gives one A.

   Sum of left-out indices	Left-out sets	Count
   0	{} (that is, A = S)	1
   1	{1}	1
   2	{2}	1
   3	{3}, {1,2}	2
   4	{4}, {1,3}	2
   5	{5}, {1,4}, {2,3}	3
   6	{6}, {1,5}, {2,4}, {1,2,3}	4
   7	{7}, {1,6}, {2,5}, {3,4}, {1,2,4}	5

3. So. The number of subsets is 1+1+1+2+2+3+4+5 = 19.

The common slip: 25. Rounding 28.8 down to 28 also counts the left-out sets with sum 8, which gives 25. A sum of 28 is below 28.8, so it fails.

The idea to keep. When a subset must have a large total, count the complements with a small total instead. Small sums are much easier to list without missing any.

Answer from the NTA final answer key. Solution by PrepWiser.