Q80 / 160 · TS EAMCET 2024 (09 May Shift 2) · Mathematics

A fair coin is tossed a fixed number of times. If the probability of getting 5 heads is equal to the probability of getting 4 heads, then the probability of getting 6 heads is

  • A
    $\frac{7}{64}$
  • B
    $\frac{9}{32}$
  • Correct answer
    C
    $\frac{21}{128}$
  • D
    $\frac{35}{256}$

Solution

Here $p(H)=\frac{1}{2}=p, q=\frac{1}{2}$ According to question, $\begin{aligned} & P(X=5)=P(X=4) \\ & \Rightarrow{ }^n C_5\left(\frac{1}{2}\right)^5 \cdot\left(\frac{1}{2}\right)^{n-5}={ }^n C_4\left(\frac{1}{2}\right)^4\left(\frac{1}{2}\right)^{n-4} \\ & \Rightarrow{ }^n C_5\left(\frac{1}{2}\right)^n={ }^n C_4\left(\frac{1}{2}\right)^n \\ & \Rightarrow{ }^n C_5={ }^n C_4 \Rightarrow n=5+4=9 \\ & \therefore P(X=6)={ }^9 C_6\left(\frac{1}{2}\right)^6\left(\frac{1}{2}\right)^3=\frac{9 \cdot 8 \cdot 7}{3 \cdot 2 \cdot 1}\left(\frac{1}{2}\right)^9=\frac{21}{128} . \end{aligned}$

Hence, the correct answer is option: 'C'.

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