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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

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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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