A fair coin is tossed a fixed number of times. If the probability of getting 7 heads is equal to probability of getting 9 heads, then the probability of getting 2 heads is :
Solution
Let the coin be tossed n-times<br><br>$P(H) = P(T) = {1 \over 2}$<br><br>P(7 heads) = $${}^n{C_7}{\left( {{1 \over 2}} \right)^{n - 7}}{\left( {{1 \over 2}} \right)^7} = {{{}^n{C_7}} \over {{2^n}}}$$<br><br>P(9 heads) = $${}^n{C_9}{\left( {{1 \over 2}} \right)^{n - 9}}{\left( {{1 \over 2}} \right)^9} = {{{}^n{C_9}} \over {{2^n}}}$$<br><br>P(7 heads) = P(9 heads)<br><br>${}^n{C_7} = {}^n{C_9} \Rightarrow n = 16$<br><br>P(2 heads) = $${}^{16}{C_2}{\left( {{1 \over 2}} \right)^{14}}{\left( {{1 \over 2}} \right)^2} = {{15 \times 8} \over {{2^{16}}}}$$<br><br>P(2 heads) $= {{15} \over {{2^{13}}}}$
About this question
Subject: Mathematics · Chapter: Probability · Topic: Classical and Axiomatic Probability
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