Q81 / 160 · TS EAMCET 2020 (14 Sep Shift 1) · Mathematics

A fair coin is tossed $K$ times so that the probability of getting four heads is equal to that of getting six heads. If the probability is maximum for getting $r$ heads then $r=$

  • A
    9
  • Correct answer
    B
    5
  • C
    8
  • D
    10

Solution

Given, $\begin{aligned} n= & K, p=\frac{1}{2}, q=\frac{1}{2} \\ & P(X=4)=P(X=6)\end{aligned}$ $\begin{aligned} \because \quad{ }^K C_4\left(\frac{1}{2}\right)^K={ }^K C_6\left(\frac{1}{2}\right)^K \Rightarrow{ }^K C_4={ }^K C_6 \\ K=4+6\left[\because{ }^n C_x={ }^n C_y \text {, then } n=x+y\right] \\ K=10 \\ P(X=r)={ }^{10} C_r\left(\frac{1}{2}\right)^{10} \end{aligned}$ Probability is maximum when $\Rightarrow r=\frac{10}{2}=5$

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

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