Let the probability of getting head for a biased coin be 1/4. It is tossed repeatedly until a head appears. Let N be the number of tosses required. If the probability that the equation 64 x²+5 Nx+1=0 has no real root is p/q, where p and q are coprime, then q-p is equal to ________.

Solution

Maths sir · 4 steps

No real root means a negative discriminant, and that limits N to 1, 2 or 3. Then add three geometric terms.

1. Condition on N. The equation has no real root when b² - 4ac < 0. Here that is 25N² - 4 × 64 × 1 < 0, so N² < 256/25 = 10.24.
2. Allowed values. N counts tosses, so it is a positive integer. N² < 10.24 allows only N = 1, 2, 3, because 4² = 16 is already too big.

   P(N = n) = (3/4)ⁿ⁻¹ × 1/4

3. Add the cases. The first head comes on toss n only after n - 1 tails. So P(N ≤ 3) = 1/4 + 3/16 + 9/64 = (16 + 12 + 9)/64 = 37/64. As a check, this equals 1 - (3/4)³, the chance that the first three tosses are not all tails.
4. So. p = 37 and q = 64 have no common factor, so q - p = 64 - 37 = 27.

The common slip: 3. Writing the discriminant as 25N² - 64 drops the 4 in 4ac. That allows only N = 1, gives probability 1/4, and makes q - p = 3.

The idea to keep. Turn the condition on the equation into a range of integer values of the random variable. For a geometric variable, P(N ≤ k) = 1 - (1-p)ᵏ, which saves you adding the terms one by one.

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