Q. 10 How many times must a man toss a fair coin so that the probability of having at least one head is more than 90^{o}/_{o} ?

Answers (1)

Let the man toss coin n times.

Probability of getting head in first toss = P

                                 P=\frac{1}{2}

                                  q=\frac{1}{2}

\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x

                               =^nC_x.\left ( \frac{1}{2} \right )^{n-x}.\left ( \frac{1}{2} \right )^x                                     

        P(getting\, atleast\, one\, head)> \frac{90}{100}

                     P(X\geq 1)> 0.9

                       1-P(X=0)> 0.9

                    1-^nC_0 \frac{1}{2^n}> 0.9

                    ^nC_0 \frac{1}{2^n}< 0.1

                        \frac{1}{2^n}< 0.1

                          \frac{1}{0.1}< 2^n

                      10< 2^n

The minimum value to satisfy the equation is 4.

The man should toss a coin 4 or more times.

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