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Question Number 24998 by anuj saini last updated on 30/Nov/17

$$\mathrm{A}\mathrm{fair}\mathrm{coin}\mathrm{is}\mathrm{tossed}100\mathrm{times}.\mathrm{The}$$$$\mathrm{probability}\mathrm{of}\mathrm{getting}\mathrm{tails}\mathrm{an}\mathrm{odd}$$$$\mathrm{number}\mathrm{of}\mathrm{times}\mathrm{is}$$

Commented by Penguin last updated on 01/Dec/17

$$attempt?$$$$P\left(\mathrm{tails}\right)=\frac{1}{2}$$$$\mathrm{from}1\mathrm{to}100,\mathrm{there}\mathrm{are}50\mathrm{even},50\mathrm{odd}\mathrm{numbers}$$$$\mathrm{so},\mathrm{there}\mathrm{is}\mathrm{an}\mathrm{equal}\mathrm{chance}\mathrm{of}\mathrm{heads}\mathrm{or}\mathrm{tails}\mathrm{and}$$$$\mathrm{odd}\mathrm{or}\mathrm{even},$$$$P(tails,odd)=\frac{1}{2}$$

Commented by prakash jain last updated on 01/Dec/17

$$\mathrm{Question}\mathrm{is}\mathrm{asking}\mathrm{for}\mathrm{a}\mathrm{probablity}$$$$\mathrm{that}\mathrm{odd}\mathrm{number}\mathrm{of}\mathrm{tails}$$$$\mathrm{you}\mathrm{are}\mathrm{correct}\mathrm{in}\mathrm{saying}\mathrm{that}$$$$\mathrm{probablity}\mathrm{of}\mathrm{odd}\mathrm{number}\mathrm{of}$$$$\mathrm{tails}=\mathrm{odd}\mathrm{number}\mathrm{of}\mathrm{heads}.$$$$\mathrm{However}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{above}\mathrm{is}\mathrm{not}$$$$\mathrm{equal}\mathrm{to}1\mathrm{since}\mathrm{that}\mathrm{are}\mathrm{cases}\mathrm{that}$$$$\mathrm{you}\mathrm{will}\mathrm{get}\mathrm{even}\mathrm{number}\mathrm{of}\mathrm{heads}$$$$\mathrm{and}\mathrm{tails}.$$

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