JAMIA 2022 — Mathematics PYQ

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Question

JAMIA 2022 — Mathematics PYQ

JAMIA | Mathematics | 2022

100 identical coins, each with probability p, of showing heads are tossed. If 0 <p < 1 and the probability of showing heads on 50 coins is equal to that of heads showing up on 51 coins, then value of p is

Choose the correct answer:

Explanation

1. Identify the Given Information

We are given:

Total number of coins ( $n$ ) = 100
$P(\text{50 heads}) = P(\text{51 heads})$
$0 < p < 1$

2. Set Up the Equation

Using the Binomial formula for $k=50$ and $k=51$ :

\binom{100}{50} p^{50} (1-p)^{100-50} = \binom{100}{51} p^{51} (1-p)^{100-51}

Simplify the powers:

\binom{100}{50} p^{50} (1-p)^{50} = \binom{100}{51} p^{51} (1-p)^{49}

3. Simplify the Terms

Divide both sides by $p^{50}$ and $(1-p)^{49}$ :

\binom{100}{50} (1-p) = \binom{100}{51} p

Now, expand the combinations using the formula $\binom{n}{r} = \frac{n!}{r!(n-r)!}$ :

\frac{100!}{50! 50!} (1-p) = \frac{100!}{51! 49!} p

Cancel out the $100!$ from both sides:

\frac{1}{50! 50!} (1-p) = \frac{1}{51! 49!} p

4. Solve for $p$

Rearrange the factorials:

\frac{51!}{50!} \cdot \frac{49!}{50!} (1-p) = p

Since $51! = 51 \times 50!$ and $50! = 50 \times 49!$ , we get:

\frac{51 \cdot 50!}{50!} \cdot \frac{49!}{50 \cdot 49!} (1-p) = p

\frac{51}{50} (1-p) = p

Multiply by 50 to clear the fraction:

51(1-p) = 50p

51 - 51p = 50p

51 = 101p

5. Final Calculation

Divide by 101:

p = \frac{51}{101}

Choose the correct answer:

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