The minimum number of times a fair coin needs to be tossed, so that the probability of getting at least two heads is at least 0.96, is

Solution:

Let the coin be tossed n times and let X-denote the number of heads inn tosses of the coin. Then,

$∴ P (X = r) =^{n} C_{r} {(\frac{1}{2})}^{r} {(\frac{1}{2})}^{n - r} =^{n} C_{r} {(\frac{1}{2})}^{n}; r = 0, 1, 2, \dots, n$

t is given that

$\begin{array}{l} P (X \geq 2) \geq 0.96 \\ \Rightarrow \sum_{r = 2}^{n} P (X = r) \geq 0.96 \end{array}$

$\begin{array}{l} \Rightarrow \sum_{r = 2}^{n}^{n} C_{r} {(\frac{1}{2})}^{n} \geq 0.96 \\ \Rightarrow \frac{1}{2^{n}} (\sum_{r = 2}^{n}^{n} C_{r}) \geq 0.96 \\ \Rightarrow \frac{1}{2^{n}} (2^{n} -^{n} C_{0} -^{n} C_{1}) \geq 0.96 \\ \Rightarrow 1 - \frac{n + 1}{2^{n}} \geq 0.96 \\ \Rightarrow (n + 1) \leq (0.04) 2^{n} \\ \Rightarrow n = 8, 9, 10, \dots \end{array}$

Hence, the least value of n is 8.

Solution:

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