Let the probability of getting head for… - JEE Main 2023 Mathematics PYQ

JEE Main 2023 — Mathematics Probability & Statistics

2023integerhard

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 $64{x}^{2}+5Nx+1=0$ has no real root is $\frac{p}{q}$ , where $p$ and $q$ are co-prime, then $q-p$ is equal to..........

Official previous-year question

Held on 11 Apr 2023 · Verified 6 Jul 2026.

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Solution

Given a quadratic equation $64{x}^{2}+5nx+1=0$ has no real roots, so

$D<0$

$\Rightarrow 25{N}^{2}-4\times 64<0$

$\Rightarrow N<\frac{16}{5}\Rightarrow N=1,2,3$

And, For a biased coin, the probability of getting head is $P=\frac{1}{4},$ and probability of getting tail will be $Q=\frac{3}{4}$

So, required probability is

$\frac{1}{4}+\frac{3}{4}\cdot \frac{1}{4}+{(\frac{3}{4})}^{2}\cdot \frac{1}{4}+{(\frac{3}{4})}^{3}\cdot \frac{1}{4}=\frac{p}{q}$

$\Rightarrow \frac{\frac{1}{4}(1-{(\frac{3}{4})}^{3})}{1-(\frac{3}{4})}=\frac{p}{q}$

$\Rightarrow 1-{(\frac{3}{4})}^{3}=\frac{p}{q}$

$\Rightarrow 1-\frac{27}{64}=\frac{p}{q}$

$\Rightarrow \frac{p}{q}=\frac{37}{64}$

$q-p=27$

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