Let `n` times a coin must be tossed so that probablity of having at least one head is more than `90%`.
`P(`At least one head`) = 1-P(`All tail`)`
`=>1-(1/2)^n > 90/100`
`=>(1/2)^n >=0.1`
Adding log base 10, to both sides,
`n(log1 - log2) > -1`
`-nlog2 > -1`
`n>1/log2=>n >1/0.3=>n>3.33`
So, coin must be tossed at least 4 times so that probablity of having at least one head is more than `90%`.