Probability XII — JEE Main · 2023 · 11 Apr 2023 · Shift 2 Question with Solution

Let the probability of getting head for a biased coin be

\frac{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 x^{2} + 5 N x + 1 = 0

has no real root is

\frac{p}{q}

, where

p

and

q

are co-prime, then

q - p

is equal to

Answer: 27

Solution

For the quadratic

64 x^{2} + 5 N x + 1 = 0

to have no real roots:

D = (5 N)^{2} - 4 (64) (1) < 0 ⟹ 25 N^{2} < 256 ⟹ N^{2} < \frac{256}{25} ⟹ N < \frac{16}{5} = 3.2

Since

N

is a positive integer,

N \in {1, 2, 3}

.

Probability of getting first head on toss

N

P (N = k) = (\frac{3}{4})^{k - 1} \cdot \frac{1}{4}

Required probability:

P (N \leq 3) = \frac{1}{4} + \frac{3}{4} \cdot \frac{1}{4} + (\frac{3}{4})^{2} \cdot \frac{1}{4} = \frac{1}{4} (1 + \frac{3}{4} + \frac{9}{16}) = \frac{1}{4} \cdot \frac{37}{16} = \frac{37}{64}

Thus

p = 37, q = 64

, and

q - p = 64 - 37 = 27

JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Integer answer