NVAMediumJEE 2023Probability Distributions

JEE Mathematics 2023 Question with Solution

Let the probability of getting a head for a biased coin be 14\frac{1}{4}. It is tossed repeatedly until a head appears. Let NN be the number of tosses required. If the probability that the equation 64x2+5Nx+1=064x^2 + 5Nx + 1 = 0 has no real root is pq\frac{p}{q}, where pp and qq are co-prime, then qpq - p is equal to _____.

Answer

Correct answer:27

Step-by-step solution

Standard Method

Given: The probability of head on each toss is 14\frac{1}{4}, so NN follows a geometric distribution with

P(N=n)=(34)n1(14),n1P(N=n)=\left(\frac{3}{4}\right)^{n-1}\left(\frac{1}{4}\right), \quad n\ge 1

Find: The value of qpq-p when the probability of the quadratic 64x2+5Nx+1=064x^2+5Nx+1=0 having no real root is pq\frac{p}{q}.

For the quadratic equation to have no real root, its discriminant must be negative:

D=(5N)24641=25N2256<0D=(5N)^2-4\cdot 64\cdot 1=25N^2-256<0

So,

25N2<25625N^2<256 N2<25625N^2<\frac{256}{25}

Hence,

N<165N<\frac{16}{5}

Since NN is a positive integer, the possible values are N=1,2,3N=1,2,3.

Now compute

P(N3)=P(N=1)+P(N=2)+P(N=3)P(N\le 3)=P(N=1)+P(N=2)+P(N=3)

Using the geometric distribution,

P(N=1)=14,P(N=2)=3414,P(N=3)=(34)214P(N=1)=\frac{1}{4}, \quad P(N=2)=\frac{3}{4}\cdot\frac{1}{4}, \quad P(N=3)=\left(\frac{3}{4}\right)^2\cdot\frac{1}{4}

Therefore,

P(N3)=14+316+964=16+12+964=3764P(N\le 3)=\frac{1}{4}+\frac{3}{16}+\frac{9}{64}=\frac{16+12+9}{64}=\frac{37}{64}

Thus pq=3764\frac{p}{q}=\frac{37}{64}. Since the fraction must be written as pq\frac{p}{q} with pp and qq coprime, we have p=37p=37 and q=64q=64. Therefore,

qp=6437=27q-p=64-37=27

So the required answer is 2727.

The solution shows an intermediate mismatch in labeling pp and qq, but the final computed value of qpq-p is correctly 2727.

Use complement form of geometric probability

Given: No real root occurs when N=1,2,3N=1,2,3. Find: The probability of this event quickly.

Instead of adding three terms separately, use

P(N3)=1P(N4)P(N\le 3)=1-P(N\ge 4)

For a geometric random variable, P(N4)P(N\ge 4) means the first three tosses are all tails:

P(N4)=(34)3=2764P(N\ge 4)=\left(\frac{3}{4}\right)^3=\frac{27}{64}

Hence,

P(N3)=12764=3764P(N\le 3)=1-\frac{27}{64}=\frac{37}{64}

So p=37,q=64p=37, q=64 and

qp=27q-p=27

Therefore, the required answer is 2727.

Common mistakes

  • A common mistake is taking the condition for no real roots as D0D\le 0 instead of D<0D<0. That includes repeated real roots, which are still real. Use strictly negative discriminant for non-real roots.

  • Students often forget that NN is the number of tosses until the first head, so NN starts from 11, not 00. Using N=0N=0 makes the geometric model incorrect.

  • Another mistake is computing P(N=3)P(N=3) as 341414\frac{3}{4}\cdot\frac{1}{4}\cdot\frac{1}{4}. The first two tosses must both be tails before the first head on the third toss, so it should be (34)214\left(\frac{3}{4}\right)^2\cdot\frac{1}{4}.

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