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# A Fair Coin is Tossed a Fixed Number of Times. If the Probability of Getting Seven Heads is Equal to that of Getting Nine Heads, the Probability of Getting Two Heads is (A) 15/28 (B) 2/15 (C) 15/213 - CBSE (Commerce) Class 12 - Mathematics

ConceptBernoulli Trials and Binomial Distribution

#### Question

A fair coin is tossed a fixed number of times. If the probability of getting seven heads is equal to that of getting nine heads, the probability of getting two heads is

• 15/28

• 2/15

• 15/213

• None of these

#### Solution

15/213
Let denote the number of heads in a fixed number of tosses of a coin .Then, is a binomial variate with parameters ,

$n \text{ and } p = \frac{1}{2}$
Given that P (X=7) =P (X = 9).
Also, p = q = 0.5
$P(X = r) = ^{n}{}{C}_r (0 . 5 )^r (0 . 5 )^{n - r} = ^{n}{}{C}_r (0 . 5 )^n$
$\therefore P(X = 7) = ^{n}{}{C}_7 (0 . 5 )^n$
$\text{ and} \ P(X = 9) =^{n}{}{C}_9 (0 . 5 )^n$
$\text{ It is given that } P(X = 7) = P(X = 9)$
$\therefore ^{n}{}{C}_7 (0 . 5 )^n =^{n}{}{C}_9 (0 . 5 )^n$
$\Rightarrow \frac{n!}{7! \left( n - 7 \right) !} = \frac{n!}{9! \left( n - 9 \right) !}$
$\Rightarrow 9 \times 8 = \left( n - 7 \right)\left( n - 8 \right)$
$\Rightarrow n^2 - 8n - 7n + 56 = 72$
$\Rightarrow n^2 - 15n - 16 = 0$
$\Rightarrow \left( n + 1 \right)\left( n - 16 \right) = 0$
$\Rightarrow n = - 1 \ or \ n = 16$
$\Rightarrow n = - 1 (\text{ Not possible as n denotes the number of tosses of a coin } )$
$\therefore n = 16$
$\text{ Hence,} \ P(X = 2) = ^{16}{}{C}_2 (0 . 5 )^{16}$
$= \frac{16 . 15}{2} \times \frac{1}{2^{16}}$
$= \frac{15}{2^{13}}$

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Solution A Fair Coin is Tossed a Fixed Number of Times. If the Probability of Getting Seven Heads is Equal to that of Getting Nine Heads, the Probability of Getting Two Heads is (A) 15/28 (B) 2/15 (C) 15/213 Concept: Bernoulli Trials and Binomial Distribution.
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