How many times must a man toss a fair coin so that the probability of having at least one head is more than 90%?

  • Option 1)

    2

  • Option 2)

    3

  • Option 3)

    4

  • Option 4)

    5

 

Answers (1)

 

Binomial Theorem on Probability -

Then

\dpi{100} P\left ( X= r \right )     or   P(r)

=\ ^{n}C_{\Upsilon }\cdot P^{r }\cdot q^{n-r}

-

 

 at least one head

 1-\frac{1}{2^{n}}> \frac{90}{100}

1-\frac{9}{10}> \frac{1}{2^{n}}

1-\frac{1}{10}> \frac{1}{2^{n}}

\therefore 2^{n} > 10

\therefore n =4 min value  option 

 


Option 1)

2

This option is incorrect

Option 2)

3

This option is incorrect

Option 3)

4

This option is correct

Option 4)

5

This option is incorrect

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